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Math 140 Quiz 2 - Summer 2004 Solution Review (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)

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Presentation on theme: "Math 140 Quiz 2 - Summer 2004 Solution Review (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)"— Presentation transcript:

1 Math 140 Quiz 2 - Summer 2004 Solution Review (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)

2 Problem 1 (4) What is the name of the shape in the Figure 1 graph show above? Parabola This was a gift!

3 Problem 2 (8) What is the name of the shape in the Figure 2 graph show above? Hyperbola Another gift!

4 Problem 3 (24) List the intercepts of the graph. Tell if the graph in Figure 1 is symmetric with respect to the x-axis, y- axis, origin, or none of these. Vertex at (0,1) is only intercept (y). Obvious y-axis symmetry

5 Problem 4 (24) Use the distance formula to find the distance between the pair of points: (7, -1); (5, -7). Let P 1 = (7, -1) & P 2 = (5, -7). Then, d(P 1, P 2 ) = [(x 1 - x 2 ) 2 + (y 1 - y 2 ) 2 ] 1/2 = [(7 - 5) 2 + (-1 - (-7)) 2 ] 1/2 = [ ] 1/2 = [40] 1/2 = (4 ) 1/2 (10 ) 1/2 = 2 (10 ) 1/2 =

6 Problem 5 (28) Write a general formula to describe the variation. A varies directly with t 2 ; A = 50 when t = 5. “A varies directly with t 2” => A = kt 2. “A = 50 when t = 5” => 50 = k(5) 2 = 25k => k = 2. So A = 2t 2.

7 Rearrange to y = (- 3/4)x + (14/4) = - (3/4)x + (7/2). Slope: m = - (3/4). Problem 6 (20) Find the slope of the line: 3x + 4y = 14.

8 Problem 7 (40) Solve the problem: The price p and the quantity x sold of a certain product obey the demand equation: p = - (1/5) x + 100, 0 < x < 500. What is the revenue when 200 units are sold? Recall revenue, R, is price times quantity or R = px. Thus, the equation to evaluate for x = 200 is R = px = [- (1/5) x + 100] x = [- (1/5) ]200 = $12000.

9 Rearrange to y = (- x - 9)/5 = - (1/5)x - 9/5. This a straight line of slope m = - 1/5 and y-axis intercept b = - 9/5. Problem 8 (72) Graph the equation and select the matching graph: - x = 5y + 9.

10 Problem 9 (40) List the intercepts for the equation: x 2 – 2x – 3 = 0. Note the factoring: x 2 – 2x – 3 = (x – 3)(x + 1). The y-axis intercept has y-coordinate that is value of the function when x = 0. This is just – 3. So intercept point is at (0, – 3). Any x-axis intercept has x-coordinate such that x 2 – 2x – 3 = (x – 3)(x + 1) = 0 => x = 3 or – 1. Thus, the x-axis intercepts are at (3, 0) and (– 1, 0).

11 Problem 10 (48) Determine if the given function is even or odd with x-axis or y-axis symmetry: f(x) = 4x 2 + x 4. f(-x) = 4(-x) 2 + (-x) 4 = 4x 2 + x 4 = f(x) Therefore, f is an even function and has y-axis symmetry.

12 Move 3 to left, square & move x 2 to left to spot circle form: x 2 + (y –3) 2 = 3 2. Thus, y is lower-half, r = 3, circle with center shifted to ( h, k ) = (0, 3). Problem 11 (60) Graph the equation and select the matching graph: y =. The y-axis intercept is at y(0) = 0.

13 Problem 12 (60) Decide whether the pair of lines is parallel, perpendicular, or neither.The figure shows the graph of two lines. Which of the pairs of equations has such a graph? Note both lines have positive slope while one has positive & one has negative y-intercept. Of answer choices, this true of only C), x - y = -2, x - y = 1. => y = x + 2, y = x - 1

14 Problem 13 (64) The graph in a figure shows a circle C. Which of the following equations has this graph? (x - (-1)) 2 + (y - 3) 2 = 4 2 Recall: (x - h) 2 + (y - k) 2 = r 2. Of answer choices, this true of only E), (x + 1) 2 + (y - 3) 2 = 16. => Note C has a radius of r = 4 and a center at: (h, k) = (-1, 3).

15 Problem 14 (76) Write an equation for the ellipse with foci at (-2, 1) and (4, 1); major axis length of 10. Note: Major axis = 2 a = 10 means a = 5. Also, foci are on line y = 1, which is parallel to x -axis, at distance 2 c = |4-(-2)| = 6. So c = 3. Thus, c 2 = a 2 - b 2 => b 2 = a 2 - c 2 = = 16 => b = 4. Center is at midpoint between foci: ( h, k ) = (1, 1). (x –h) 2 + (y –k) 2 = 1 becomes answer D), _ a 2 b 2 (x –1) 2 + (y –1) 2 = 1. See next slide for graph. _ 25 16

16 Problem 14 continued Write an equation for the ellipse with foci at (-2, 1) and (4, 1); major axis length of 10. 2c Center: ( h, k ) = (1,1) 2a = 10 2b

17 Eliminate B), & C) choices by finding f(1) is not on graph; D) choice is straight line not present. For A): f(x) =. Problem 15 (56) Select the matching function to the graph. This works since domain is x -1 > 0 as shown. Also testing points like x = 1, 2, 3,... works. Finally, note image is like a rotated & shifted parabola: x = y x f(1) = 0, f(2) = 1, f(5) = 2.

18 Eliminate A), C), and D), choices by finding f(2) or is not on graph. For B): f(x) = |x +2| means when x + 2 > 0, f(x) = |x +2| = x +2, which is a line starting at (-2, 0) with m = 1 as shown in right part. Problem 16 (40) Select the matching function to the graph. When x +2 < 0, f(x) = |x +2| = -(x +2) which is a line starting at (-2, 0) with m = -1 as shown in graph’s left part. Thus, function is: f(x) = |x +2|. x + 2 0

19 Problem 17 (48) Find the requested function, f – g, with f(x) = 9x - 8 and g(x) = 3x – 7. (f – g)(x) = f(x) - g(x) = 9x - 8 – (3x – 7) = 9x - 3x = 6x - 1

20 Problem 18 (36) Find the requested function, ( f g)(-2), with f(x) = x + 4 and g(x) = 3x x - 6. (f g)(x) = f(x)g(x) = (x + 4)(3x x - 6) ( f g)(-2) = [(-2) + 4][3(-2) (-2) – 6] = 2(-24) = -48

21 Problem 19 (40) Find the requested function, g o f, with f(x) = (x – 9)/10 and g(x) = 10x + 9. (g o f )(x) = g(f(x)) = 10[(x – 9)/10] + 9 = (x – 9) + 9 = x

22 Factor –2 & complete square: y = -2[x 2 + x + (½) 2 - (½) 2 ] –7 = -2(x + ½) 2 – 6½ This an inverted compressed parabola with vertex shifted to ( h, k ) = (-½, - 6½). Problem 20 (68) Graph the equation and select the matching graph: y = -2x 2 - 2x – 7. The y-axis intercept is at y(0) = –7.

23 Problem 21 (88) From the Lines Figure, determine the equation of the line going through the point (0, 1) that is perpendicular to the two parallel lines shown. From Problem 12 the slope of both lines is m 1 = 1. Thus, slope of the  line is m 2 = -1/m 1 = -1. Pt. Slope: y - y 1 = m 2 (x – x 1 ). y - 1 = -1(x – 0) => y = - x + 1

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