Questions and Answers #1 Polynomial Graphs

Questions and Answers Sheet 1 Polynomial Graphs Question #1 Graph the polynomial function. 𝑓(𝑥) = −𝑥 4 Answer: Step 1 𝑓(𝑥) = −𝑥 4 The leading coefficient is -1 (negative) and the degree if 4 (even) Therefore 𝑓(𝑥) ⇒ −, as 𝑥 ⇒ ∞ and 𝑓(𝑥) ⇒ −∞, as 𝑥 ⇒ −∞ Step 2 To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior. Question #2 Graph the polynomial by transforming an appropriate graph of the form 𝑦 = 𝑥 𝑛 . Show clearly all x- and y-intercepts. 𝑃(𝑥) = 81 − (𝑥 − 3)4 Answer: Step 1 The parent function of 𝑃(𝑥) = 81 − (𝑥 − 3)4 is 𝑦 = 𝑥 4 , which passed through the points (−2,16), (−1,1), (0,0), (1,1), and (2,16). 𝑃(𝑥) = −(𝑥 − 3)4 + 81 is the graph of 𝑦 = 𝑥 4 reflected across the x-axis and then translated left 3 units right and up 81 units. Reflecting the points on 𝑦 = 𝑥 4 accross the x-axis given the points (−2, −16), (−1, −1), (0,0), (1, −1), and (2, −16) . Translating these points right 3 units and up 81 units then gives the points (1,65), (2,80), (3,81), (4,80), and (5,65). The x-intercept of P(x) is when 𝑃(𝑥) = 0: 𝑃(𝑥) = 81 − (𝑥 − 3)4 Given function. 0 = 81 − (𝑥 − 3)4 Substitute 𝑃(𝑥) = 0 (𝑥 − 3)4 = 81 Add (𝑥 − 3)4 4 𝑥 − 3 = ± √81 Take the 4th root of both sides. 𝑥 − 3 = ±3 Simplify 𝑥 = 3 ± 3 Add 3 on both sides The y-intercept of P(x) are then 𝑥 = 3 − 3 = 0 and 𝑥 = 3 + 3 = 6 The y-intercept is when 𝑥 = 0 . From finding the x-intercept, we know 𝑦 = 0 when 𝑥 = 0 so the yintercept is at (0,0). Plot the points and then connect them with a smoth curve. Label the coordinates of the intercepts in your graph: Answer: See the explanation for the graph. To make the graph, reflect the graph of 𝑦 = 𝑥 4 across the xaxis and then translate right 3 units and up 81 units. Label the intercepts at (0,0) and (6,0) Question #3 Graph the polynomial by transforming an appropriate graph of the form 𝑦 = 𝑥 𝑛 . Show clearly all x- and y-intercepts. 𝑃(𝑥) = −𝑥 3 + 64 Answer: Step 1 The parent function of 𝑃(𝑥) = −𝑥 3 + 64 is 𝑦 = 𝑥 3 , which passes through the points (−2, −8), (−1, −1), (0,0), (1,1) and (2,8) . 𝑃(𝑥) = −𝑥 3 + 64 is the graph of 𝑦 = 𝑥 3 reflected across the x-axis and then translated up 64 units. Reflecting the points on 𝑦 = 𝑥 3 across the x-axis gives (−2,8), (−1,1), (0,0), (1, −1) and (2, −8). Translating these points up 64 units then gives the points (−2,72), (−1,65), (0,64), (1,63), and (2,56). The x-intercept of 𝑃(𝑥) is when 𝑃(𝑥) = 0: 𝑃(𝑥) = −𝑥 3 + 64 Given function. 0 = −𝑥 3 + 64 Substitute 𝑃(𝑥) = 0 𝑥 3 = 64 Add 𝑥 3 on both sides. 3 𝑥 = √64 Cube root both sides. 𝑥 = 4 Simplify The x-intercept of P(x) is then the point (4,0). The y-intercept is when 𝑥 = 0 which we already know is at the point (0,64). Plot the points, including the x-intercept, and then connect them with a smooth curve. Label the coordinates of the intercepts in your graph: Question #4 Graph each polynomial function. 𝑓(𝑥) = 𝑥 3 + 3𝑥 2 − 4𝑥 − 2 Answer: Step 1 Use a graphing calculator to graph 𝑓(𝑥) = 𝑥 3 + 3𝑥 2 − 4𝑥 − 2 The x-intercept are (−3.9,0), (−0.4,0) and (1.29,0) The y -intercept is (0,-2) Question #5 Graph each polynomial function. 𝑓(𝑥) = (𝑥 − 1)4 Answer: Step 1 𝑓(𝑥) = (𝑥 − 1)4 To find the x-intercept solve 𝑓(𝑥) = 0 (𝑥 − 1)4 = 0 The given equation 𝑥 − 1 = 0𝑥 𝑎 = 0 so 𝑥 = 0 𝑥 = 1 Solve The x-intercept is (1,0) To find the y-intercept evaluate f(0) 𝑓(𝑥) = (𝑥 − 1)4 The given equation 𝑓(0) = (0 − 1)4 Substitute 0 for x =1 Simplify The y-intercept is (0,1) Plot the points (−1,16), (2,1), and (3,16) See the graph below Question #6 Use a graphing calculator to examine the graphs of the following functions. Determine ⋅⋅ the degree of the polynomial ⋅⋅ the leading coefficient ⋅⋅ the end behavior of the polynomial function ⋅⋅ the maximum number of zeros ⋅⋅ the maximum number of turning points (relative maxima and minima) 1 a. 𝑓(𝑥) = 𝑥 2 + 2𝑥 2 b. 𝑓(𝑥) = 𝑥(𝑥 + 1)2 (𝑥 − 1)2 c. 𝑓(𝑥) = 𝑥 3 − 𝑥 2 + 31𝑥 + 40 Answer: Step 1 a.Degree:2(highest-degree monomial is Leading coefficient: 1 2 1 2 𝑥 2 ). (coefficient of highest-degree monomial) End behavior: 𝑦 ⇒ ∞ as 𝑥 ⇒ ±∞ (even-degree polynomial with positive leading coefficient) Maximum number of zeros: 2(same as degree) Maximum number of turning points: 1(one less than degree) Step 2 b. Degree: 5(product of 5 linear factors) Leading coefficient: 1(product of coefficient of x terms) End behavior: 𝑦 ⇒ ∞ as 𝑥 ⇒ ∞; 𝑦 ⇒ −∞ as ⇒ −∞ (odd-degree polynomial with positive leading coefficient) Maximum number of zeros: 5(same as degree) Maximum number of turning points: 4 (one less than degree) Step 3 c. Degree: 3 (degree of highest-degree monomial 𝑥 3 ) Leading coefficient: 1(coefficient of highest-degree monomial) End behavior: 𝑦 ⇒ ∞ as 𝑥 ⇒ ∞; 𝑦 ⇒ −∞ as 𝑥 ⇒ −∞ (odd-degree polynomial with positive leading coefficient) Maximum number of zeros: 3 (same as degree) Maximum number of turning points: 2 (one less than degree) Question #7 Investigate the change in the graph of a sinusoidal function of the form and = sin 𝑥 or and = cos 𝑥 when multiplied by a polynomial function. Use a graphing calculator to sketch the graphs of and = 𝑥 2 , and = −𝑥 2 and = 𝑥 2 sin 𝑥 on the same coordinate plane, on the interval [−20,20]. Answer: Question #8 Suppose f has period 2 and 𝑓(𝑥) = 𝑥 for 0 ≤ 𝑥 ≤; 2 . Find the fourth-degree Fourier polynomial and graph it on 0 ≤ 𝑥 ≤; 2. Answer: Step 1 The given function 𝑓(𝑥) = 𝑥, 0 ≤ 𝑥 < 2 𝑏 𝑏 Since we know the general expression of Fourier series for f on [− , ] is 2 2 𝑎0 + ∑∞ 𝑘=1 𝑎𝑘 cos ( 2𝜋𝑘𝑥 𝑏 )+ ∑∞ 𝑘=1 𝑏𝑘 sin ( 2𝜋𝑘𝑥 𝑏 Where constant expressed as 1 𝑏 ) 𝑏 2 𝑎0 = ∫2𝑏 𝑓(𝑥)𝑑𝑥 , 𝑎𝑘 = ∫2𝑏 𝑓(𝑥) cos(𝑘𝑥) 𝑑𝑥 , for 𝑘 > 0 𝑏 𝑏 − − 2 𝑏 2 2 𝑏 −𝑏 2 2 𝑏𝑘 = ∫ 𝑓(𝑥) sin(𝑘𝑥) 𝑑𝑥 , for 𝑘 > 0 Now the value of constant 1 2 1 𝑥2 2 | ] 2 2 0 𝑎0 = [∫0 𝑥𝑑𝑥 ] = [ 2 =1 Furthermore 4 2 𝑎1 = [∫0 𝑥 cos 𝜋 𝑥𝑑𝑥] = 2 [( 𝑥 sin 𝜋𝑥 2 𝜋 + cos 𝜋𝑥 𝜋2 ) |20 ] = 0 And 4 2 −𝑥 cos 𝜋 𝑥 sin 𝜋 𝑥 2 4 𝑏1 = [∫ 𝑥 sin 𝜋 𝑥𝑑𝑥] = 2 [( + ) |0 ] = − 2 2 0 𝜋 𝜋 𝜋 Therefore the fourier polynomial of degree 1 is given by 4 𝑓(𝑥) ≈ 1 − sin 𝜋 𝑥 𝜋 Step 2 Now similarly other coefficients 4 2 𝑎2 = [∫0 𝑥 cos 2 𝜋𝑥𝑑𝑥] = 2 [( 2 𝑥 sin 2𝜋𝑥 2𝜋 + cos 2𝜋𝑥 4𝜋2 ) |20 ] = 0 And 4 2 −𝑥 cos 2 𝜋𝑥 sin 2 𝜋𝑥 2 2 𝑏2 = [∫ 𝑥 sin 2 𝜋𝑥𝑑𝑥] = 2 [( + ) |0 ] = − 2 2 0 2𝜋 4𝜋 𝜋 4 2 𝑥 sin 3 𝜋𝑥 cos 3 𝜋𝑥 2 𝑎3 = [∫ 𝑥 cos 3 𝜋𝑥𝑑𝑥] = 2 [( + ) |0 ] = 0 2 0 3𝜋 9𝜋 2 And 4 2 −𝑥 cos 3 𝜋𝑥 sin 3 𝜋𝑥 2 4 𝑏3 = [∫ 𝑥 sin 3 𝜋𝑥𝑑𝑥] = 2 [( + ) |0 ] = − 2 2 0 3𝜋 9𝜋 3𝜋 4 2 𝑥 sin 4 𝜋𝑥 cos 4 𝜋𝑥 2 𝑎4 = [∫ 𝑥 cos 4 𝜋𝑥𝑑𝑥] = 2 [( + ) |0 ] = 0 2 0 4𝜋 16𝜋 2 And 4 1 − x cos 4 x sin 4 x 2 1 b4 = [  x sin 4 xdx] = 2[( + |0 ] = − 2 0 2 4 16  Now the fFourier polynomial of degree 4 is 4 2 4 𝜋 𝜋 3𝜋 𝑓(𝑥) ≈ 1 − sin 𝜋 𝑥 − sin 𝜋 𝑥 − Step 3 1 sin 3 𝜋𝑥 − sin 4 𝜋𝑥.. 𝜋 Question #9 Graph the polynomial 𝑃(𝑥) = −(𝑥 + 2)3 + 27 , showing clearly all x- and y- intercepts. Answer: Work as shown below, follow the steps: 𝑃(𝑥) = −(𝑥 + 2)3 + 27 Graph this function by using transformations of the graph of the parent function 𝑓(𝑥) = 𝑥 3 : The function P can be written in terms of the parent function f as: 𝑃(𝑥) = −𝑓(𝑥 + 2) + 27 So the transformations we have to apply to the graph of f to get the graph of P are: -Horizontal translation 2 units left. -Reflection through the x-axis. Vertical translation 27 units upward. See the graphs of these two functions in the following picture: Answer: The transformations we have to apply to the graph of f to get the graph of P are: horizontal translation 2 units left, reflection throught the x-axis, and vertical translation 27 units upward. Intercepts: 𝑥 = 1 and 𝑦 = 19 Question #10 Graph the polynomial by transforming an appropriate graph of the form 𝑦 = 𝑥 𝑛 . Show clearly all x- and y-intercepts. 𝑃(𝑥) = 2(𝑥 + 1)4 − 32 Answer: The parent function of 𝑃(𝑥) = 2(𝑥 + 1)4 − 32 is 𝑦 = 𝑥 4 , which passes through the points (−2,16), (−1,1), (0,0), (1,1), (2,16). 𝑃(𝑥) = 2(𝑥 + 1)4 − 32 is the graph of 𝑦 = 𝑥 4 vertically stratched by a factor of 2 and then translated left 1 unit and down 32 units. Vertically stretching the points on 𝑦 = 𝑥 4 by a factor of 2 gives the points (−2,32), (−1,2), (0,0), (1,2), (2,32). Translating these points left 1 unit and down 16 units then gives the points (−3,0), (−2, −30), (−1, −32), (0, −30), (1,0) The x-intercept of P(x) is when 𝑃(𝑥) = 0 , we which know is at the points (−3,0), (1,0) . The y-intercept is when 𝑥 = 0 which we already know is at the point (0,-30) . Plot the points and then connect them with a smooth curve. Label the coordinates of the intercept in your graph: