Questions and Answers #1 Linear Equations and Graphs

Questions and Answers Sheet 1 Linear Equations and Graphs Question #1 What is the graph of 𝑓(𝑥) = 𝑥 2 ? Answer: The graph of 𝑓(𝑥) = 𝑥 2 is called a ''Parabola.' It looks like this: One of the ways to graph this is to use plug in a few x-values and get an idea of the shape. Since the x values keep getting squared, there is an exponential increase on either side of the y-axis. You can see this by plugging in a few values: When 𝑥 = 0, 𝑓(𝑥) = 0 𝑥 = 1, 𝑓(𝑥) = 12 = 1 𝑥 = 2, 𝑓(𝑥) = 22 = 4 𝑥 = 3, 𝑓(𝑥) = 32 = 9 𝑥 = 4, 𝑓(𝑥) = 42 = 16 The same holds true for negative x-values to the left of the y-axis since a negative value squared is positive. For example, 𝑥 = −1, 𝑓(𝑥) = (−1)2 = −1 × −1 = 1 𝑥 = 2, 𝑓(𝑥) = (−2)2 = −2 × −2 = 4 Question #2 1 Find the linear approximation of the function 𝑔(𝑥) = (1 + 𝑥)3 at a=0 and use it to approximate the 1 1 numbers (0.95)3 and (1.1)3 . Answer: Question #3 Consider the case when the constant A=4. Plot the graph of 𝑦 = Answer: 4 𝑥 Question #4 find the area of the largest rectangle that can be inscribed in the ellipse Answer: 𝑥2 𝑎2 + 𝑦2 𝑏2 =1 Question #5 Find the area enclosed by the curve 𝑥 = 𝑡 2 − 2𝑡, 𝑦 = √1𝑡 and the y-axis. Answer: Question #6 Find the area of the region enclosed by one loop of the curve 𝑟 = sin 1 2𝜃. Find the area of the region that lies inside the first curve outside the second curve. 𝑟 = 14 cos 𝜃 , 𝑟 = 7 Answer: Question #7 Find the exact length of the curve. 𝑦 = ln(sec 𝑥) , 0 ≤ 𝑥 ≤ Answer: 𝑦 = ln(sec 𝑥) we can write the arc length formula as follows: 𝑑𝑦 2 𝑑𝑥 𝑑𝑥 𝑎 differentiate (1) with respect to x 𝑏 𝐿 = ∫ √1 + 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = sec 𝑥 tan 𝑥 sec 𝑥 = tan 𝑥 𝜋 𝐿 = ∫04 √1 + (𝑡𝑎𝑛𝑥)2 𝑑𝑥 𝜋 4 𝜋 = ∫04 √1 + tan2 𝑥 𝑑𝑥 𝜋 = ∫04 sec 𝑥 𝑑𝑥 𝜋 = [ln|sec 𝑥 + tan 𝑥|]04 = ln(√2 + 1) − ln(1 + 0) = ln(√2 + 1) Question #8 Find equations of both lines through the point (2, -3) that are tangent to the parabola 𝑦 = 𝑥 2 + 𝑥. Answer: Note that all points on the parabola 𝑦 = 𝑥 2 + 𝑥 are of the form (𝑥, 𝑥 2 + 𝑥) Assume that the tangent at (𝑎, 𝑎2 + 𝑎) passes through (2, -3) We know that the slope of tangent at any point is the derivative at that point. Differentiate 𝑦 = 𝑥 2 + 𝑥, To get 𝑑𝑦 𝑑𝑥 = 2𝑥 + 1 Therefore, the slope of tangent at (𝑎, 𝑎2 + 𝑎) 𝑖𝑠 2𝑎 + 1 ⇒ (1) Since the tangent passes through the points (𝑎, 𝑎2 + 𝑎) and (2. -3). We can write the slope of the tangent as (𝑎2 +𝑎)−(−3) 𝑎−2 = 𝑎2 +𝑎+3 𝑎−2 ⇒ (2) Using (1) and (2), we can write 2 𝑎2 +𝑎+3 𝑎−2 = 2𝑎 + 1 𝑎 + 𝑎 + 3 = (2𝑎 + 1)(𝑎 − 2) 𝑎2 + 𝑎 + 3 = 2𝑎2 + 𝑎 − 4𝑎 − 2 𝑎2 − 4𝑎 − 5 = 0 (𝑎 + 1)(𝑎 − 5) = 0 ⇒ 𝑎 = −1 and a = 5 Therefore, there are two tangents that pass through (2, -3) Finding the equation of the tangent corresponding to a = -1 Note that the slope of the tangent is 2a + 1 = 2(-1) + 1 = -1 Since the tangent. passes through (2, -3). Therefore, the equation of the tangent is 𝑦−(−3) 𝑥−2 = −1 y + 3 = -(x - 2) x+y+1=0 Finding the equation of the tangent corresponding to a = 5 Note that the slope of the tangent is 2a + 1 = 2(5) + 1 = 11 Since the tangent. passes through (2, -3). Therefore, the equation of the tangent is 𝑦−(−3) 𝑥−2 = 11 𝑦 + 3 = 11(𝑥 − 2) 𝑦 = 11𝑥 − 25 Question #9 A triangles has virtices of (0,0), (0,4), and (2,0). Find new vertices if: A. (𝑥, 𝑦) → (𝑥 − 4 + 3) B. (𝑥, 𝑦) → (−𝑥, 𝑦) C. (𝑥, 𝑦) → (−𝑦, 𝑥) Answer: A. Using the rule (𝑥, 𝑦) → (𝑥 − 4, 𝑦 + 3) (0,0) → (0 − 4,0 + 3) = (−4,3) (0,4) → (0 − 4,4 + 3) = (−4,7) (2,0) → (2 − 4,0 + 3) = (−2,3) B. (𝑥, 𝑦) → (−𝑥, 𝑦) (0,0) → (0,0) (0,4) → (0,4) (2,0) → (−2,0) C. (𝑥, 𝑦) → (−𝑦, 𝑥) (0,0) → (0,0) (0,4) → (−4,0) (2,0) → (0,2) Question #10 𝑦 = 1/2𝑥 − 9 Find the line that's perpendicular to it and passes through (7,10) Answer: First, find the slope using the formula: y=mx+b Substitute y=1/2x-9 m=1/2 Perpendicular lines have negative reciprocal slopes, therefore the needed line will have a slope of -2. Point slope form: ( ) y− y = m x−x 1 1 Substitute given coordinates: y-10=-2(x-7) y-10=-2x+14 y=-2x+24 Question #11 yˆ = 2 x + 7 x=3 y=9 Find the residual of the point Answer: Given: yˆ = 2 x + 7 (3;9) In this instance the residual is y − yˆ Substitute y=9 to yˆ = 2 x + 7 to find ŷ : yˆ = 2 ( 3) + 7 =13 Subtract to find the residual: y − yˆ = 9 − 13 =-4 Question #12 Given 𝑅(2,1), 𝑆(2,3), 𝑇(6,3), 𝑈(6,4), 𝑉(8,4) Compare the slope of segment RT and TV Answer: Segment RT: RS R − S 2 − 2 +1− 3 −2 1 = = = = S −T 2 − 6 + 3 − 3 −4 2 ST Segment TV: TU T − U 6−6+3− 4 −1 1 = = = = 6 − 8 + 4 − 4 −2 2 UV U − V Since RS 1 TU 1 = and = , the answer is: ST 2 UV 2 The slope segment of segment RT is equal to the slope segment of TV. Question #13 Given: A ( −2, −2 ) , M ( −2,3) , AM = MB Find B Answer: Since AM = MB , M is the midpoint. Let A be  x, y  , B be  x, y  and M be  x, y  . Use the 1  2  m   1  2  m formula for the midpoint: x+ x  y + y   x, y  =  1 2  ,  1 2  m    m   2   2    x+ x 1 2 =x m 2 −2 + x 2 = −2 2 −2 + x = −4 2 x = −2 2 y+ y =y 2 m −2 + y 2 =3 2 −2 + y = 6 1 2 2 y =8 2 B(-2,8) Question #14 Is 4𝑥 2 + 2𝑦 = 6 a linear equation? Give a form of a linear equation. Answer: No, it's not. The form of a linear equation: az+by=c Since x in the given equation is raised to the power of 2, it's not linear, it's quadratic. Question #15 y=36/x Find y, if x is: 1. 0.5 2. 1 3. 2 4. 3 5. 4 Answer: y=36/x 1. y=36/0.5 y=72 2. y=36/1 y=36 3. y=36/2 y=18 4. y=36/3 y=12 5. y=36/4 y=9 Question #16 8x-2y-5=11 Solve for y Answer: 8x-2y-5=11 8x-5=11+2y 2y=8x-16 y=4x-8 Question #17 Find the linearization L(x) of the function at a. 𝑓(𝑥) = √𝑥, 𝑎 = 4 Answer: 𝑓(𝑥) = √𝑥 𝑓(4) = √4 = 2 1 𝑓 ′ (𝑥) = 𝑓 ′ (4) = 2√𝑥 1 2√4 = 1 4 The linearization L(x) of f at a=4: 𝐿(𝑥) = 𝑓(𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) 1 𝐿(𝑥) = 2 + (𝑥 − 4) 1 4 𝐿(𝑥) = 𝑥 + 1 4 Results: 1 𝐿(𝑥) = 𝑥 + 1 4 Question #18 Sketch the graph of a function that satisfies all of the given conditions. f'(0)=f'(0)=f'(4)=0 f'(x)>0 if x<0 or 24 f''(x)>0 if 13 Answer: Step 1 Correct choice is the last one that is D. All the reasons are shown below in graph. Question #19 Graph the linear equations. Then use the graph to estimate how many solutions the equations share. 𝑦 = 4𝑥 + 9 4𝑥 − 𝑦 = 1 Answer: Step 1 You can graph the given equations by graphing their intercepts. Start with the first equation. To find the xintercept, substitute y in the equation for 0 and solve the equation that you get for x. y=4x+9 0=4x+9 -9=4x 9 − 𝑥 4 Step 2 List the x-intercept with both its coordinates. 9 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 (− , 0) 4 Step 3 To find the y-interceprt, substitute x in the same equation for 0 and solve the equation that you get for y. y=4x+9 𝑦 =4×0+9 y=0+9 y=9 Step 4 List the y-intercept with both its coordinates. 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡(0,9) Step 5 Now work with the other equation. To find the x-intercept, substitute y in the equation for 0 and solve the equation that you get for x. 4x-y=1 4x-0=1 4x=1 1 𝑥= 4 Step 6 List the x-intercept with both its coordinates. 1 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 ( , 0) 4 Step 7 To find the y-intercept, substitute x in the same equation for 0 and solve the equation that you get for y. 4x-y=1 4×0−𝑦 =1 0-y=1 -y=1 y=-1 Step 8 List the y-intercept with both its coordinates. 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡(0, −1) Step 9 Graph both equations by graphing their intercepts. Step 10 The two lines are parallel, so they never intersect and thus there is no solution. Question #20 Find the points on the curve 𝑦 = 2𝑥 3 + 3𝑥 2 − 12𝑥 + 1 where the tangent is horizontal. Answer: We are looking for a solution to this equation: y'(x)=0 because the tangent is horizontal when the derivative is zero. So, y'(x)=0 (2𝑥 3 + 3𝑥 2 − 12𝑥 + 1)′ = 0 6𝑥 2 + 6𝑥 − 12 = 0 𝑥2 + 𝑥 − 2 = 0 Now we need to find a solution to this equation: 𝑥2 + 𝑥 − 2 = 0 This is a quadratic equation so the solution is given with this formula: 𝑥1,2 = −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 We have 𝑥1,2 = 𝑥1,2 = 𝑥1 = −1±√12 −4⋅1⋅(−2) 2⋅1 −1±√9 2 −1+3 2 𝑥1 = 1 𝑥2 = −1−3 2 𝑥2 = 2 So the given curve has horizontal tangent when x=1,-2, so the corresponding points are (1,-6) and (-2,21) Result: (1,-6), (-2,21)