Precalculus Review Booklet - Topic 27

Topics • (27) Graphing; graphs of common functions XXVII. Graphing; graphs of common functions A graph is a pictorial presentation of n-ples (i.e., pairs, triples, quadruples, etc.) of points which satisfy a function or relation. In the case of functions or relations involving only two variables (e.g., x and y), graphs may be presented on the Cartesian plane. The Cartesian plane is the perpendicular intersection of two real-number lines at their origins. Upward movement represents positive y-movement and rightward movement represents positive x-movement. (One can imagine the Cartesian plane’s generalization to three-dimensional space, four-dimensional space, and n-dimensional space.) Functions have an independent variable and a dependent variable. The independent variable varies among the function’s domain. The dependent variable varies among the function’s range; its freedom to vary among the range is dependent on the function’s assignments. The letter x usually represents single-variable functions’ independent variables and y usually represents their dependent variables. Relations do not have independent and dependent variables. Lines Let’s consider the graphs of lines in the Cartesian plane. A general algebraic representation of a line y = mx + b. (A line is a function. One may express this as f(x) = mx + b.) m represents the slope of a line. The slope of a line is the signed (e.g., + or –) amount by which a graph moves vertically for a movement of one positive unit on the x-axis. More generally, slope is the ratio of vertical movement to horizontal movement. (For example, if a line moves downward by 5 for rightward movement of 1, its slope is -5/1 = -5. The constancy of the slope of a line, though, gives us that this line will move upward 20 for a leftward movement of 4. We have 20/-4 = -5, the slope.) b is the y-intercept, the y-value at which the graph crosses the y-axis; i.e., the y-value when x = 0. For example, the line y = 3x + 7 moves up 3 units for every rightward movement of 1 unit, and it crosses the y-axis at y = 7. Since the slope of a line is constant, one needs to locate only two points on a line’s graph to determine the whole graph. One can use the y-intercept as one of these two points. Then one can solve for the x-intercept by asking what x-value makes y = 0. In the case of y = 3x + 7, the x-intercept is the x-value such that 3x + 7 = 0. By manipulating this equation, we get x = -7/3. To graph y = 3x + 7, we plot our two known points, (0,7) and (-7/3,0), and draw a line between and through them. Both x and y can take any value in the real numbers. That is, the graph extends infinitely in the negative and positive x-directions and infinitely in the negative and positive y-directions. The presentation y = mx + b is called the slope-intercept form of the line because one can determine easily from this expression the slope and y-intercept of the line. Only one line passes through a particular point with a particular slope. In other words, a line can be determined fully from these two pieces of information (a slope and a point). If we are asked to plot the graph of a line through (2,5) with slope 4, we might begin by expressing this line in point-slope form: 4 = (y-5)/(x-2). (Recall that slope is vertical movement divided by horizontal movement.) We interpret point-slope form, in this example, as stating that 4 is the slope of the line between (2,5) and every other point (x,y) on the line. By manipulating the equation 4 = (y-5)/(x-2), we can rewrite this line in slope-intercept form: y = 4x – 3. 2 Parabolas Equations with positive integral exponents and highest exponent 2 are called quadratic equations. Their graphs are parabolas: u-like shapes whose arms extend infinitely in the negative and positive x-directions. y = x2 is the simplest quadratic equation. The following are a few points of its graph: (0,0), (1,1), (-1,1), (2,4), (-2,4), (3,9), (-3,9). Now consider y = x2 + c (c0). The following are a few points of its graph: (0,c), (1,1+c), (-1,1+c), (2,4+c), (-2,4+c), (3,9+c), (-3,9+c). Note that the parabolic shapes of y = x2 and y = x2 + c (c0) are the same. The two equation’s graphs differ in only one way: the latter’s graph is shifted up or down by |c| from the former’s situation in the plane. Now note that y = x2 and y = (x-b)2 have the same parabolic shapes. The latter equation’s graph is shifted right or left by |b| from the former’s situation in the plane. Let’s consider the graphs of the equations y = x2 and y = ax2 (a0). (0,0), (1,a), (-1,a), (2,4a), (-2,4a), (3,9a), (-3,9a) are a few points of the graph of y = ax2 (a0). Neither the graph of y = x2 nor the graph of y = ax2 is shifted up or down or right or left from the other’s position in the plane. The multiplier a may do several things, though, to the graph of y = x2. A negative multiplier will flip the parabola over the x-axis, causing it to open downward. A multiplier whose absolute value is less than 1 will slow the growth of the function’s magnitude. A multiplier whose absolute value is greater than 1 will quicken the growth of the function’s magnitude. Parabolas have a vertical line of symmetry. For example, the y-axis is the line of symmetry of the graph of y = x2. The vertical line x = 3 is the line of symmetry of the graph of y = (x-3)2. To find a parabola’s line of symmetry, complete the square of a given quadratic equation to arrive at an equation in the form f(x) = a(x-b)2 + c. The line of symmetry is the vertical line x = b. The turning point of a parabola is the point at which it reaches a maximum or minimum. The line of symmetry passes directly through the turning point. Therefore, we may find the least or greatest y-value achieved by the parabola by evaluating f(x) = a(x-b)2 + c at x = b. f(b) always will equal c since (b-b)2 = 0. That is, the least or greatest value of f(x) = a(x-b)2 + c is the value c. A discussion of the nature of complex roots Equations with positive integral exponents and highest exponent 3 are called cubic equations. Recall that the Fundamental Theorem of Algebra gives us that a cubic equation has three roots among the complex numbers. (The real numbers, remember, are a subset of the complex numbers). Complex, non-real roots of polynomial equations come in conjugate pairs. If one root is (a+bi), it will be accompanied by the root (a–bi). The following presentation of ideas will prove this. The expression (x–(5+i))(x–(5–i)) = 0, which has roots x = 5+i and x = 5–i, distributes and simplifies in the following manner: x2 – (5–i)x – (5+i)x + (5+i)(5–i) = x2 – 5x + ix – 5x – ix + (25 + 5i – 5i – i2) = x2 –10x + (25 – -1) = x2 – 10x + 26 = 0. The resulting equation, x2 – 10x + 26 = 0, has only real coefficients. That (5+i) and (5–i) are conjugates causes the terms containing i1 (x1-terms and scalar terms) to cancel each other additively. Then that i2 = -1, an integer, allows the final expression to be free of imaginary numbers. This example shows how distribution and simplification clears imaginary numbers from quadratic equations whose roots are complex conjugates. We know from the quadratic formula, though, that a quadratic equations’ non-real roots – if it has any – always will be complex conjugates. The quadratic formula reads: x = [-b  (b2 – 4ac)½]/2a. The  symbol guarantees that non-real roots will be complex conjugates. Now note that the highest exponent in a polynomial equation dictates its general shape. The highest exponent dictates global behavior because, given large numbers as input, the lead 3 monomial in a polynomial (i.e., that with highest exponent) produces much larger numbers than the non-lead monomials. For example, 10004 = 1 trillion, 10003 = 1 billion, and 10002 = 1 million. The first is one thousand times as great as the second and one million times as great as the third. Now, that odd exponents preserve negativeness and positiveness [e.g., (-2)3 = -8 and (+3)3 = +27] but even exponents produce only positiveness [e.g., (-2)2 = +4 and (-3)2 = +9] allows us to classify – from a detail-blurring perspective – polynomial functions into two groups: those with odd highest exponent and those with even highest exponent. The graphs of the former stretch vertically from negative infinity to positive infinity. Those of the latter extend vertically to either positive infinity or negative infinity but never both. If polynomial functions have as their domain the entire real continuum, their graphs will have no breaks. Therefore, such polynomial functions with odd highest exponent will intersect the x-axis at some point. We do not have this guarantee with polynomial functions with even highest exponent, but we do know that they will have an absolute minimum (if the lead coefficient is positive) or absolute maximum value (if the lead coefficient is negative). These classes and observations allow us great insight into the common problem of finding roots of polynomial functions. The nature of polynomial functions with odd highest exponent is such that we always can remove from it a real-termed factor. After one such factoring, we are left to find the roots of an expression with even highest exponent. (For example, suppose we are asked to find all of the roots of x3 – 8 = 0. We saw when we discussed long division that x3 – 8 = (x – 2)(x2 + 2x + 4). Therefore, we know that x = 2 is a root; and to continue finding roots, we need to find those of x2 + 2x + 4 = 0, an even polynomial function.) A polynomial function with even highest exponent will not cross the x-axis if its minimum sits above the x-axis or its maximum sits below the x-axis. If an even polynomial function never crosses the x-axis, all of its roots will be non-real. A polynomial function whose greatest exponent is even may cross the x-axis, but the preceding allows us to say that every non-real root is the root of an even polynomial function. The Fundamental Theorem of Algebra gives us that an even polynomial function has an even number of roots in the complex numbers. Evenness is a structural necessity for our proof because the extreme case – that in which an even polynomial function has only non-real roots – requires an even number of roots to accommodate a remainderless pairing of non-real roots. We are poised now to prove in a formal manner that non-real roots come in conjugate pairs. Initially, I attempted to show that non-real roots come in conjugate pairs by employing induction. Proof by induction establishes as true a basic case and seeks to build a ladder from this basic case up to the next case and all future cases. For example, one might prove by induction that the sum 1 + 2 + 3 +…+ n equals ½n(n+1). This statement certainly holds for the case in which n = 1, the basic case. In the next step, we assume that the statement holds in the first n cases and we utilize this assumption to show that the statement holds in the n+1st case. If we show this, we will know that the n+1st case holds if the nth case holds. However, our previous establishment (our in-duction) of the basic case transforms this conditionality into certainty: the 1st case gives the 2nd case, the 2nd gives the 3rd, the 3rd gives the 4th, etc. Induction suits our proof regarding non-real roots because we need to show that a statement holds for quadratics, cubics, quartics, quintics, etc.; in other words, for integrally infinitely many cases. Already we reduced the set of polynomials with which we need to deal to the set of those of even degree. Also, we established the truth of our statement for the case in which n = 2, the least even number. As I built a proof by induction, though, its layers of complexity, mostly algebraic, began to pile on and obscure the simple drive of the proof. Such aim-clouding – straying seemingly far from the central track of a proof – often is necessary in establishing mathematical truths, but I started to consider seriously that another path might lead more clearly and quickly to the desired demonstration. The following presents another path. 4 Our goal is to show that if a+bi [or a–bi] is a root of any polynomial, then a–bi [or a+bi] is another root of that polynomial. Consider the case of the general quadratic equation, keeping in mind that we need to show that our assertion holds for all polynomials. If a+bi is a root of x2 + c1x + c0, then we know that (a+bi)2 + c1(a+bi) + c0 = 0. The full distribution and simplification of the left side of this equation yields that a2 + 2abi – b2 + c1a + c1bi + c0 = 0. If we segregate real and imaginary terms, we can express the left side of the equation in the standard form of a complex number [r + ri: r, i = (-1)½]. After such segregation, we have that (a2 – b2 + c1a + c0) + (2ab + c1b)i = R1 + R2i = 0. Our segregation is sensible because addition among real numbers cannot yield imaginary terms and addition among imaginary terms cannot yield real numbers. This additive incapacity tells us that both R1 and R2 equal zero. Now we want to test whether a–bi is a root or not. To do this, we change b to -b wherever b occurs. R1 remains unchanged because b2 = (-b)2. R2 changes, though, to -R2; but -R2 = 0 because R2 = 0. Therefore, a–bi is a root of x2 + c1x + c0. Recall from our discussion above that all polynomials of odd degree have at least one real root and that, consequently, we may confine our non-real-root-investigative attention to polynomials of even degree. Accordingly, we consider next the case of the quartic. If a+bi is a root of x4 + c3x3 + c2x2 + c1x + c0, then – after full distribution, simplification, and segregation – we know that (a4 + b4 + c3a3 – 6a2b2 – 3c3ab2 + c2a2 – c2b2 + c1a + c0) + (4a3b – 4ab3 – c3b3 + 3c3a2b + 2c2ab + c1b)i = R1 + R2i = 0. Substituting -b for b, we find again that R1 remains unchanged, R2 becomes -R2, and, therefore, that a–bi is a root. We would achieve the same result were we to continue upward through the polynomials of even degree. Note the 4-cylical nature of i: i = i, i2 = -1, i3 = i2·i = (-1)i = -i, i4 = i2·i2 = (-1)(-1) = 1, i5 = i4·i = 1·i = i, i6 = i4·i2 = 1(-1) = -1, etc. The four spokes of the cycle are i, -1, -i, and 1. From the preceding, one can see that in is an integer if n is even and an imaginary number if n is odd. Therefore, in the full distribution of the expression (a+bi)n [and, by a natural extension, the full distribution of (a+bi)n + cn–1(a+bi)n–1 + cn–2(a+bi)n–2 +…+ c1x + c0 (ci )], we will find imaginary terms only where we find b raised to an odd power. Consequently, in all of the b-containing real terms in the full distribution of (a+bi)n [and (a+bi)n + cn–1(a+bi)n–1 + cn–2(a+bi)n–2 +…+ c1x + c0 (ci )], we will find b raised to an even power. Thus, when we substitute -b for b, all of the real terms remain unchanged because (-b)n = bn if n is even, and all of the imaginary terms negate because (-b)n = -bn if n is odd. Hence, whenever a+bi is the root of a polynomial with real coefficients, a–bi is another root. Cubic functions From the Fundamental Theorem of Algebra and the discussion above, we know that a cubic function has either three real roots or one real root and a conjugate pair of non-real roots. Cubic functions must cross the x-axis once, but they never cross only twice. Recognizing that cubic functions’ behavior is so restricted, and understanding why, will aid one’s graphing and root-seeking endeavors. Although the cubic function y = (x–2)(x–2)(x–1) appears to cross the x-axis exactly twice (at x = 1 and x = 2), it crosses only once (at x = 1). The value x = 2 is a double root. At a double root, a function touches the x-axis and promptly turns around – never crossing the x-axis. Contrast the case of the triple root. The value x = 0 is a triple root of 3 y = x . At x = 0, the function touches the x-axis, turns around once, and turns again – all at one point. Thus, at a triple root, one will see a single x-axis-crossing. A cubic function which has distinct roots (i.e., which has neither a triple nor double root) will have one upward-facing banking (or rounding, or turning) and one downward-facing banking. In contrast, quadratic functions have one upward- or downward-facing banking. For all polynomial functions, the 5 addition of a degree of variability creates the potential for an additional banking. For example, a twenty-eighth-degree polynomial with twenty-eight distinct roots has twenty-seven bankings. Quartic functions By the fundamental Theorem of Algebra and our discussion of the nature of complex roots, we know that quartic functions have zero, two, or four complex, non-real roots. Therefore, they may cross the x-axis four times, two times, or zero times. The graphs of many (but not all) quartic functions have the shape of a rounded W (opening either upward or downward). A quartic function with four distinct real roots will cross the x-axis four times and exhibit the shape of a rounded W. The following passages present other scenarios one may encounter among quartic functions. The function y = x4 – x2 = x2(x2 – 1) has one double root, x = 0. At the point (0,0), the graph of this function, having approached from below the x-axis, touches (but does not cross) that axis, and immediately turns downward. Despite having only three roots, the graph retains the shape of a W. The graph of the function y = x2(x – 1)2, which has two double roots, x = 0 and x = 1, retains the shape of a W, but it never crosses the x-axis. The graph touches and immediately turns at the points (0,0) and (1,0). A quartic function with a triple root (e.g., y = x(x – 1)3) will have a graph that crosses the x-axis twice but fails to have the shape of a W, having lost one of its bumps to the triple root. A quartic function with a quadruple root has only one bump and never crosses the x-axis. The examples above involve quartic functions with only real roots. If a given quartic function has four non-real roots, one can manipulate the standard expression of the function (i.e., y = c4x4 + c3x3 + c2x2 + c1x + c0) to facilitate the sketching of a graph. Remove c0 (which will be positive in this case) from the standard expression of the given quartic function and graph the resulting function. After removing c0, one always can factor an x from the expression that remains; then one needs to factor only a cubic expression. Cubic expressions have either three real roots or only one. In the former case, one has four real roots to guide one’s sketch. In sketching, plot points between the outermost roots and their immediate neighbors to learn whether there is or is not a significant difference in the depth of the W’s two local minima. After sketching the graph, raise the graph by c0 to produce a graph of the original function. In the latter case (that in which the cubic expression has only one real root), lower the altered function more to force the existence of real roots. Also, if a given quartic function has only two non-real roots, try lowering the function to force the existence of real roots. After graphing the lowered function, raise it to produce a graph of the given function. Quintic, sextic, septic, etc., functions Graphing quintic, sextic, septic, etc., functions is generally similar to graphing functions of lower degree. One needs to find roots. (Recall the Fundamental Theorem of Algebra and that complex, non-real roots come in conjugate pairs.). One needs to determine how multiple roots, if a given function has any, affect a graph’s behavior near and at the x-axis. Also, one needs to determine the shapes of graphs in regions where local minima are above the x-axis and local maxima are below it (i.e., in cases in which functions have non-real roots). (See the final paragraph of the subsection on quartic functions.) Reciprocals of linear expressions (and their vertical transpositions) Consider the reciprocals of linear expressions (i.e., expressions mx + b). For example, the function y = 1/x = x-1, the reciprocal of the linear expression x, has positive and negative branches, both of which are asymptotic to the x- and y-axes. (Asymptotes are functions to 6 which more complex functions collapse as x or y becomes infinite or infinitesimal. An asymptote can be crossed at finite values, but the asymptotes of x-1 are never crossed.) As lines extend to positive infinity in one x-direction and negative infinity in the other, their reciprocals become small positive numbers in one x-direction and small negative numbers in the other. That one cannot divide a quantity by zero prevents one from finding the reciprocal of a line where it crosses the x-axis. In the region about an x-axis-crossing, a line achieves arbitrarily small yvalues (positive and negative). Therefore, the reciprocal of a line grows arbitrarily large (positively or negatively) in that region. If a line crosses the x-axis at x = r, the asymptotes of its reciprocal function will be the vertical line x = r and the x-axis (i.e., the horizontal line y = 0). One may add a constant c to the reciprocal of a line. The resulting equation, y = [1/(mx + b)] + c, will not be the reciprocal of a line (if c is nonzero). By adding c, one shifts upward or downward the horizontal asymptote of y = 1/(mx + b) but does not change its vertical asymptote. Circles A circle is the set of all points distant from a single point (the center) by a particular positive real number. Since the x- and y-axes in the Cartesian plane are perpendicular to each other, the Pythagorean Theorem gives us the symbolic expression for a circle. For example, x2 + y2 = r2 is the set of all points (x,y) which create a right triangle whose right angle is coincident with the intersection of the x- and y-axes and whose hypotenuse has length r. All of these points (x,y) are distant by r from the origin, the center of this circle. Recall that the introduction of b into the parabola y = x2 to create y = (x-b)2 shifts the graph of y = x2 to the right or left. With similar alterations, we may move the graph of a circle up or down or to the right or left. We begin with x2 + y2 = r2 and introduce nonzero b and c to produce the new circle (x-b)2 + (y-c)2 = r2 with center (b,c). Adjusting the values b, c, and r allows us to produce any circle. Note that since circles fail the vertical line test, they are relations and not functions. However, one may divide a circle into portions expressible as functions. Draw a horizontal diameter through a circle, dividing it into an upper half and a lower half. If we so divide x2 + y2 = r2, two functions result: y = +(r2 – x2)½ and y = -(r2 – x2)½. The graphs of these two functions join to become one circle. Ellipses Consider the equation x2 + 4y2 = 25. If we rewrite this equation as (x2 + y2) + 3y2 = 25, we see that it falls short of being the equation of a circle. The graph of x2 + 4y2 = 25 is similar, though, to that of a circle – it is an ellipse. The following presents rigorously (but not in mathematical symbols) the relationship between circles and ellipses. Assume we are given a circle. Among the infinitely many choices, we choose one diameter and inflate it. Inflation – rather than radical or apical extension, for example – increases the magnitude of every possible segment of this diameter by the same proportion at the same time. We can associate every segment of the pre-inflationary diameter with a piece of the circle by extending straight lines perpendicularly from a segment’s endpoints on both sides of the diameter. Each segment’s perpendicular lines will mark off two circular segments when they intersect the circle – one circular segment for each semicircle created by our introduction of a diameter into the circle. As the diameter inflates, every diametric segment’s two associated circular segments must inflate so that their endpoints maintain the same distance from the diameter. This is diameter-dependent inflation. Independent inflation would give us a larger circle. Diameter-dependent inflation unidirectionally stretches the circle into a (non-circular) ellipse. I placed non-circular in parentheses because circles are ellipses, too – just as squares are rectangles, too. 7 The diameter along which we stretched our circle is the major axis of resulting ellipses. We call the line running from one side of the ellipse to the other and perpendicularly bisecting our inflated diameter, the minor axis. Note that the minor axis would remain unchanged if we were to run our inflation backward; i.e., if we were to deflate the ellipse down to its original, circular state. However, if we were to continue the deflation past circularity, the minor axis would become the major axis. A major axis in a non-circular ellipse is so called because it is longer than the other axis. The equation a12(x-b)2 + a22(y-c)2 = k2 exhibits the standard presentation of ellipses whose major axes are parallel to the x- or y-axis. The major axis of an ellipse can have any orientation in the plane, but we deal here with only the cases represented by a12(x-b)2 + a22(y-c)2 = k2. To produce ellipses whose major axes are not parallel to the x- or y-axis, one can convert ellipses conforming to a12(x-b)2 + a22(y-c)2 = k2 into polar coordinates [Let x = rcosθ and y = rsinθ. Radius r is positive and measured always from the origin.] and turn them counterclockwise by β radians by substituting (θ–β) for θ wherever θ occurs. Use angle-addition formulae to expand cos(θ–β) and sin(θ–β) and then convert back to Cartesian coordinates. The major and minor axes of ellipses conforming to a12(x-b)2 + a22(y-c)2 = k2 intersect at (b,c). If a12 < a22, the major axis will be parallel to the x-axis. If a12 > a22, the major axis will be parallel to the y-axis. Despite that (b,c) is the intersection of the axes, it cannot be called the center of the ellipse. Ellipses do not have centers. They have a two-centric system of foci: for every point p on the ellipse, the sum of (1) the distance from focus1 to p and (2) the distance from focus2 to p, is the same. Circles are ellipses, recall. In circles, the two centers have the same position. For every point p on a circle, the sum of (1) the distance from the center to p and (2) the distance from p back to the center is the same. This sum is 2r. One can graph of an ellipse without employing its foci. To sketch the graph of an ellipse, find the points at which the major and minor axes intersect with the ellipse. Consider again x2 + 4y2 = 25. When y = 0, x2 = 25; that is, x = 5. When x = 0, 4y2 = 25; that is, y = 5/2. Therefore, the major axis intersects the ellipse at (5,0) and (-5,0) and the minor axis intersects the ellipse at (0,5/2) and (0,-5/2). Drawing a smooth and vertically and horizontally symmetrical curve through these four points suffices to sketch the ellipse. If one wants to produce a more accurate drawing, one can plug nonzero x’s or y’s into x2 + 4y2 = 25 to find other points on the ellipse. Despite that foci do not play a role in the sketching of a symbolically expressed ellipse, one may be interested to know where they lie within a given ellipse. The idea that one can produce an ellipse by diameter-dependently inflating a circle suggests that an ellipse’s foci are on the diameter that guided the inflation; i.e., on an ellipse’s major axis. Before we confirm the location of foci, let us be sure that ellipses have a two-centric system of foci that operates as I have stated. One might test my statement about the character of ellipses’ foci in the following way. Gather a non-square rectangular piece of paper, a segment of string whose length is that of the shorter dimension of the paper, a bit of tape, and a pen or pencil. Position the paper so that its longer dimension runs horizontally. On the line parallel to and halfway between the top and bottom of the paper, mark two points so that both have the same distance from the vertical edges and the distance between them is shorter than the length of string already cut. Tape the ends of the string to these points. With the pen or pencil, push the string to tautness. Always holding the string taut and with pen or pencil to paper, move the pen or pencil in a 360-degree arc. The shape that emerges should look elliptical. We will need, however, to use algebra to prove that this process produces an ellipse. 8 Let us impose a Cartesian plane onto our rectangular piece of paper, positioning the xaxis so that it runs horizontally and so that the x- and y-axes cross at the intersection of the rectangle’s diagonal lines. The foci we marked earlier are on the x-axis and are equidistant from the origin. Let us say each has distance d from the origin. The coordinates of the foci, therefore, are (-d,0) and (d,0). Let us say the length of the string is l. We wish to create an equation in x and y that gives the relationship between the x’s and y’s in all of the (x,y)-pairs comprising the shape that emerges in the process described above. When we hold the string taut with a pen or pencil, we create straight lines, which are easy to measure analytically with the distance formula. The distance formula is a direct application of the Pythagorean Theorem. In our present case, we need to find the distance between focus1 [the point (-d,0)] and some point (x,y) and the distance between focus2 [the point (d,0)] and (x,y). Each distance is regarded as the hypotenuse of a right triangle. The first distance is [(x – -d)2 + (y – 0)2]½ and the second is [(x – d)2 + (y – 0)2]½. The sum of these two distances, for all x and y, is l, the length of the string. Therefore, we have that [(x + d)2 + y2]½ + [(x – d)2 + y2]½ = l. Let us manipulate this equation to put it in its simplest form. We begin by maximally executing distributive multiplication within each bracketed expression: [(x2 + 2dx+ d2 + y2]½ + [(x2 – 2dx + d2 + y2]½ = l. Move the first bracketed expression to the right side of the equation: [(x2 – 2dx + d2 + y2]½ = l – [(x2 + 2dx+ d2 + y2]½. Now square both sides of the equation: x2 – 2dx + d2 + y2 = l2 + x2 + 2dx+ d2 + y2 – 2l[(x2 + 2dx+ d2 + y2]½. Cancel all terms common to both sides of the equation, leaving: -2dx = l2 + 2dx – 2l[(x2 + 2dx+ d2 + y2]½. Now move – 2dx to the right side of the equation and -2l[(x2 + 2dx+ d2 + y2]½ to the left side: 2l[(x2 + 2dx+ d2 + y2]½ = l2 + 4dx. Square both sides of the equation: 4l2(x2 + 2dx+ d2 + y2) = l4 + 16d2x2 + 8l2dx. On the left side of the equation, distribute 4l2 over the four terms in parentheses: 4l2x2 + 8l2dx + 4l2d2 + 4l2y2 = l4 + 16d2x2 + 8l2dx. Cancel the one term (8l2) common to both sides: 4l2x2 + 4l2d2 + 4l2y2 = l4 + 16d2x2. Now gather all variable-containing terms on the left side of the equation and all constant terms on the right side (in a particular iteration of this shapedrawing process, d and l are constant): 4l2x2 – 16d2x2+ 4l2y2 = l4 – 4l2d2. Factor x2 from the first two terms on the left side of the equation: (4l2 – 16d2)x2 + 4l2y2 = l4 – 4l2d2. This equation conforms to the standard presentation of an ellipse whose major axis is parallel to the x- or yaxis! In this case, the major axis is parallel to the x-axis because 4l2 – 16d2 is less than 4l2. The intersection of the major and minor axes is the point (b,c) = (0,0), which tells us that the major axis is incident with the x-axis. Now we know that the foci of this ellipse – and every ellipse – are on its major axis and equidistant from the intersection of its major and minor axes. We have not developed yet a way to determine focal coordinates from the symbolic expression of an ellipse. Assume we are given an ellipse whose major axis is parallel to the x- or y-axis; i.e., one which can be made to have the form of a12(x-b)2 + a22(y-c)2 = k2. (Recall that we deal here with only ellipses that can be made to have this form.) By noting whether a12 is greater or less than a22, determine whether the major axis is parallel to the x- or y-axis. Find the coordinates of the endpoints of the major and minor axes by letting y = c and solving for x (two values) and letting x = b and solving for y (two values). Give the name M to one-half of the length of the major axis and the name m to one-half of the length of the minor axis. Now imagine a string slackly connecting the two foci – which we know are along the major axis, equidistant from the intersection of the major and minor axes. Give the name d to the distance from each focus to that intersection. When the string running between the foci is pulled taut, it creates two sides of a triangle (the triangle’s other side is the straight line segment connecting the two foci). As the taut string revolves about the intersection of the major and minor axes – its third vertex tracing an ellipse – the shape of this triangle changes continuously. At the moment at which the third vertex meets one (or the other) endpoint of the major axis, both string-created sides of the triangle are incident with the major axis. The triangle at this moment 9 is flat – its area is zero. At this instant, we can see that the length of the string is (M + d) + (M – d) = 2M, the full length of the major axis. We can use this length to determine the coordinates of the foci. Allow the taut string to revolve another 90˚. The third vertex is now incident with one of the endpoints of the minor axis, and the string-created triangle is isosceles. The height of this isosceles triangle is m. We can divide an isosceles along its altitude into congruent right triangles. The two legs of this triangle are d (unknown) and m (known), and its hypotenuse is M (known). By the Pythagorean theorem, we know that d2 + m2 = M2. Therefore, d2 = M2 – m2 and d equals the positive (recall that d is a distance) square root of M2 – m2. One can now give the coordinates of the foci: (bd, c) if the major axis is horizontal, and (b,cd) if it is vertical. Hyperbolas Consider the equations x2 – y2 = r2 and x2 – 9y2 = k2. The graphs of these equations are hyperbolas. The minus signs in the left sides of these equations distinguish them from equations of circles and ellipses. One might regard a hyperbola loosely as the product of the following process: divide a circle or ellipse in halves along its minor (or major) axis; turn the resulting arms outward along the natural extensions of its major (or minor) axis – maintaining the arms’ equidistance from that axis and pivoting at the endpoints of that axis; now allow these arms to extend infinitely upward and downward as they extend infinitely left and right. An ellipse (a12(x-b)2 + a22(y-c)2 = k2) and one of its correlative hyperbolas (a12(x-b)2 – a22(y-c)2 = k2 or a22(y-c)2 – a12(x-b)2 = k2) have only two common points, the endpoints of the major or minor axes. This characterization of hyperbolas is loose because we have not stated exactly how to alter the shape of the arms once they have been turned outward. Of course, an hyperbola’s equation strictly governs this alteration. Let’s consider an example: x2 – 4y2 = 25. To create this hyperbola from its correlative ellipse, x2 + 4y2 = 25, we divide the ellipse into halves along its minor axis and turn the arms outward along extensions of its major axis, pivoting at the endpoints of the major axis. After this turning, the left arms extend left and the right arms extend right, but these finite portions of the hyperbola’s arms do not conform to the hyperbola’s equation. To conform, the arms’ curvature must adjust a bit (all change in the same way), setting a course for their infinite extensions. These extensions are guided by linear functions – asymptotes. Let us study the hyperbola’s equation to determine its asymptotes. The standard presentation, x2 – 4y2 = 25, may be changed to x2 = 4y2 + 25. The constant term ensures that x2 never equals 4y2, but when x and y are large, x2 and 4y2 overwhelm 25. As x and y move toward negative and positive infinity, x2 and 4y2 begin to dwarf 25 and become arbitrarily close to each other. (Only that they must remain unequal keeps them apart.) This arbitrariness allows us to say that x2 and 4y2 have a kind of equality: asymptotic equality. We write x2 ~ 4y2. Therefore, x ~ 2y; y = ½x are the equations of our asymptotes. These lines create a large X whose center is at the origin. (The hyperbola’s equation is (x–0)2 – 4(y–0)2 = 25). From a perspective sufficiently encompassing that the hyperbola’s details are blurred, the hyperbola and this X are identical. However, from an ordinary perspective, focused on the center of the X, we see that the hyperbola’s left and right wings never intersect. The hyperbola’s correlative ellipse has exactly sufficient room to rest between the wings. The hyperbolas we have considered to this point are correlates of ellipses whose major axes are parallel to the x- or y-axis. As we turn an ellipse about the intersection of its major and minor axes (its unofficial center), we turn its correlative hyperbolas (there are two for every ellipse: e.g., x2 + 4y2 = 25 spawns x2 – 4y2 = 25 and 4y2 – x2 = 25). An example of an hyperbola whose situation in the plane differs from those of hyperbolas that can conform to the standard presentations a12(x-b)2 – a22(y-c)2 = k2 and a22(y-c)2 – a12(x-b)2 = k2 is the function y = 1/x. Let us standardize y = 1/x to prove that it is an hyperbola. A sketch of y = 1/x suggests that it is a 10 standard hyperbola shifted 45° counterclockwise from the position exhibited by x2 – 4y2 = 25. (Also, one may say that it is a standard hyperbola shifted 45° clockwise from the position exhibited by 4y2 – x2 = 25.) Write y = 1/x as xy = 1 and express the latter in polar coordinates: (rcosθ)(rsinθ) = 1. To undo a counterclockwise shift, we shift clockwise. Recall that 45° = π/4 radians. In (rcosθ)(rsinθ) = 1, substitute (θ – -π/4) = (θ + π/4) where θ occurs: [rcos(θ + π/4)][rsin(θ + π/4)] = 1. Expand these expressions by employing the angle-addition formulae for sine and cosine: r[cosθcos(π/4) – sinθsin(π/4)]·r[sinθcos(π/4) + cosθsin(π/4)] = 1. Now maximally execute distributive multiplication: r2sinθcosθcos2(π/4) + r2cos2θsin(π/4)cos(π/4) – r2sin2θsin(π/4)cos(π/4) – r2sinθcosθsin2(π/4) = 1. Recall that cos(π/4) = sin(π/4) = ½½ and substitute numbers for non-variable trigonometric expressions: ½r2sinθcosθ + ½r2cos2θ – ½r2sin2θ – ½r2sinθcosθ = 1. Simplify the left side of the equation: ½r2cos2θ – ½r2sin2θ = 1. Multiply both sides of the equation by 2: r2cos2θ – r2sin2θ = 2. Now express the equation in x and y: x2 – y2 = 2. We now have an hyperbola in one of the two standard forms. If we had shifted y = 1/x counterclockwise by 45° = π/4 radians, our final equation would have been y2 – x2 = 2, which exhibits the other standard form. Logarithms and exponential functions The domains of logarithms are always the entirety of the positive real numbers and their ranges are always the entirety of the real numbers. Therefore, graphs of logarithms extend vertically from negative to positive infinity and extend horizontally from the y-axis, which they never touch and to which they are asymptotic in their lower reaches, to positive infinity. A logarithm can differ from another in only one manner: it may have a different base. However, distinct logarithms logax and logbx (a ≠ b) have great similarity. One’s graph is a y-axis-dependent inflation or deflation of the other. In other words, the two differ by only a constant multiplier: specifically, we may write that logax = (logab)logbx. (See section VII to review how one can move from one base to another.) That all logarithms have shape-similarity and that all cross the x-axis at x = 1 allow us to sketch them easily. Also, that any two logarithms are proportional to each other allows us to express all logarithms with one base and an appropriate multiplier. For reasons we will discuss in the sections ahead, e = 2.71… is the universal base in mathematics. The logarithm is unique in that it grows more slowly than any other kind of function. (The function log(logbx), for example, grows more slowly than logbx grows, but I do not regard log(logbx) as an essential peer of logbx because it is constructed from the logarithm.) When x is large, logbx will differ negligibly or almost negligibly from logb(x–1). In other words, at a logarithm graph’s far-right reaches, rightward movement produces relatively little vertical movement. At the far-right, the logarithmic curve has the appearance of a horizontal line, but it never reaches perfect horizontality. Recall that a function which passes the horizontal-line test (i.e., each y-value corresponds to only one x-value) can be reflected over the line y = x to produce a new function. This new function is the original’s inverse function. The logarithm’s inverse function, as we saw in section VI, is basex, the exponential function. For example, the function y = 2x is the reflection over y = x of the function y = log2x. Therefore, log2x and 2x have the same (rather than similar) shape but different orientation in the plane. The exponential function sits above the x-axis (never touching it) and always crosses the y-axis at y = 1.) The transcendental number e = 2.71… is the universal base for exponential functions, too. If one would like to express y = 2x with the base e = 2.71…, one needs to determine k such that ek = 2; then y = 2x = (ek)x = ekx. Also, that the logarithm grows more slowly than any other kind of function, gives us (according to reflection over y = x) that the exponential function grows more quickly than any other kind of function. (The function b^(bx), for example, grows more 11 quickly than bx grows, but I do not regard b^(bx) as in essential parity with bx because it is constructed from the exponential function.) The sine, cosine, and tangent functions Recall that sinθ = cos(/2–θ). The graphs of the sine and cosine functions have the same (rather than similar) shape, but each is shifted horizontally from the other’s situation in the plane by π/2 radians. The graphs of sine and cosine are perfectly undulating waves. They exhibit the purest form of two-dimensional wave activity. With proper adjustments to amplitude, period, and horizontal situation in the plane, one can add together sine waves (or cosine waves) to produce any kind of wave within a closed stretch of the x-axis. Unaltered sine and cosine waves move between y = -1 and y = 1 and exhibit one wave (i.e., one peak and one valley) every 2π radians (360º). The domain of the sine and cosine functions is the set of all real numbers. Recall that tanθ = sinθ/cosθ. We know, therefore, that the domain of the tangent function is the set of all real numbers other than those that would force division by zero. In other words, tangent’s domain includes all real numbers other than θ such that cosθ = 0. The angles θ between 0 and 2π such that cosθ = 0 are π/2 and 3π/2. We need a way to express succinctly the set of all θ throughout the real numbers such that cosθ = 0. The expression {θ: cosθ = 0} (read this as “the set of θ in the real numbers such that cosθ = 0”) certainly suffices but does not imply a method by which one can compute any such angle. If one looks to x-stretches other than the one from 0 to 2π, including negative stretches, one will begin to notice that angles at which cosine equals zero are odd multiples of π/2. The full set is {…-7π/2, -5π/2, -3π/2, -π/2, π/2, 3π/2, 5π/2, 7π/2, …}. We can express this set non-elliptically: {θ = (2k+1)π/2: kZ}. Now we must determine how the graph of tangent behaves at such θ. Our discussion of division by zero in section XIII showed that it creates infinite behavior – either negatively or positively infinite or both. Consider tangent at θ = π/2, for example. At θ = π/2, sine reaches its apex, 1. Therefore, within small neighborhoods about θ = π/2, sine remains positive. As we move from θ = π/2 to the right (i.e., toward π), cosine moves from zero toward negative-one through arbitrarily small values. We have then that tangent at θ that are close to but to the right of π/2 will be negative and great. On the left side of θ = π/2, cosine is arbitrarily small and positive. Therefore, at θ that are close to but to the left of π/2, tangent will be great and positive. We see thus that tangent exhibits both negatively and positively infinite behavior at θ = π/2. Now consider θ = 3π/2. At θ = 3π/2, sine reaches its nadir, -1. Within small neighborhoods about θ = 3π/2, sine remains negative. As θ passes from left to right through 3π/2, cosine passes through first arbitrarily small negative values, then zero, and then arbitrarily small positive values. Tangent therefore exhibits positively infinite behavior to the left of 3π/2 [a negative number divided by a negative number yields a positive number] and negatively infinite behavior to the right of 3π/2. Again, we see infinite behavior of two kinds at one point. Moreover, in both cases we have examined, we see positively infinite behavior on the left and negatively infinite behavior on the right. Note that every pair of consecutive angles θ = (2k+1)π/2 and θ = (2k+3)π/2 contains one at which sinθ = 1 and one at which sinθ = -1. Hence, by considering θ = π/2 and θ = 3π/2, we covered all [of the two] cases. Now, by observing that sine – and therefore tangent – has zero value at all points exactly halfway between consecutive θ in {θ = (2k+1)π/2: kZ}, we come to see that within every interval whose open-endpoints are a pair of consecutive angles in {θ = (2k+1)π/2: kZ}, tangent exhibits the same graph. (Tangent’s period is π rather than sine’s and cosine’s 2π). From left to right, this graph rushes up from great negative values, crosses the x-axis at the interval’s midpoint, and surges up to great positive values. 12 Homework Problem Set 5 Topics • (27) Graphing; graphs of common functions Please refrain from using a calculator to solve or complete any of the following. 27a. Graph the line y = 5x + 2. What is the slope of this line? What are its x- and yintercepts? 27b. Graph the line 9x – 7y + 15 = 0. (Rearrange into slope-intercept form; i.e., the form of y = mx + b.) 27c. Are the solutions to x2 + 5 = 0 real numbers? Graph y = x2 + 5. 27d. Graph of y = x2 – 14x + 45. 27e. Graph y = x3 + 10x2 + 3x. (Factor.) 27f. Graph y = x4 – 13x2 + 36. (This quartic function has a quadratic structure.) 27g. Graph y = x5 + x3 + x – 3. (Use long division to factor.) 27h. Graph y = 1/x, y = 1/(x + 5), and y = (x + 6)/(x + 5). (Try altering the third function by subtracting and adding a number from the numerator.) 27i. Graph of (x + 6)2 + (y – 2)2 = 169. 27j. Graph 4(x – 3)2 + 9(y – 5)2 = 144. 27k. Graph 5(x – 3)2 + 5(y + 7)2 = 125. 27l. Graph x2 – y2 = 2 and xy = 1. 27m. Graph 2(y – 5)2 – 3(x – 3)2 = 9. (Begin by graphing the correlative ellipse.) 27n. Graph y = log2x and y = log10x on the same plane – in different colors if possible. 27oa. Now graph y = 2x and y = 8x on the same plane – in different colors if possible. 27ob. Express y = 8x as an exponential function with base 2. 27p. Graph y = sinθ and y = sin(2θ) on the same plane – in different colors if possible. 27q. Graph one period of the secant function. (Recall that secθ is the reciprocal of cosθ.) Solve the following word problems without using a calculator. Try graphing the function or functions implied in each problem. 27r. A private jet leaves Brainard Airport (Hartford, CT) at 10:00 a.m. bound for LAX in Los Angeles. Its cruising speed is 200 m.p.h. At 3:00 p.m. on the same day, a large passenger jet leaves Bradley Airport (Windsor Locks, CT – close to Hartford) for LAX and settles quickly into a 450-m.p.h. cruising speed. If the two jets take the same flight path, at what time will the large one pass the small one? 27sa. A gardener has 50 feet of fencing with which to enclose a rectangular area for a garden. What are the dimensions of the largest such garden he can enclose? 27sb. Now assume the gardener has 60 feet of fencing to enclose a triangular garden? [This problem implies a function of two variables. Try finding A = f(a,b) = the area of the triangle as a function of the lengths of two sides of the triangle.] 27t. What is the volume of the largest rectangular, square-bottomed box one can cover with 100 cm2 of newspaper? 13 27u. Two desk clocks are plugged into an AC outlet and are set for noon within seconds of each other at noon on July 1, 2005. At noon on the next day, clock1 reads 12:00 and clock2 reads 11:59. At noon on July 3, clock1 reads 12:00 and clock2 reads 11:56. At noon on July 4, clock2 reads 11:51, and at noon on July 5, clock2 reads 11:44. For months, clock1 steadily displays the correct time, but clock2 slows down continuously and maximally smoothly (i.e., bumpiness in graph is minimal) according to the pattern established above. Find the next day and time at which the two clocks will give the same time, assuming that they give twenty-four-hour time; e.g., 3:00 p.m. reads 15:00. 27v. Suppose we have a water faucet controlled by a circular dial with a rotational range of 360 = 2 radians. Counterclockwise motion increases the flow of water, which exits the faucet at 3α cubic centimeters/second: α is the angle in radians through which the faucet has moved from its zero position, three o’clock. Assume the dial is at six o’clock when we stick a cylindrical cup with radius 8cm and height 15cm beneath the faucet. If we begin to lessen the flow of water at the moment the cup is touched by falling water and we restrict ourselves to (1) a constant rate of angular movement of the dial and (2) the condition that at the moment the cup is full, the dial returns to zero position; how many seconds pass from the first drop’s exit to the last drop’s touchdown? 27x. After construction of an in-ground swimming pool was aborted for lack of funds, Diana was left with a hole in the shape of a right circular frustum. The bottom of the frustum has diameter 20 feet, the top has diameter 25 feet, and the depth of the hole is 12 feet. If Diana is able to refill the hole at the rate of 3 cubic feet per minute, how many hours will Diana need to spend shoveling dirt into the hole? (Try graphing hours of shoveling done (on x-axis) versus volume of remaining hole (y-axis). To determine the volume of a frustum, regard it as a truncated cone, find the volume of the full cone and that of the absent portion of the full cone (the absent portion is a smaller and similar cone), and subtract the latter from the former.)