Chapter 2 Linear and Quadratic Functions Section 2.1 7. y-intercept = 2 − 3 , we see that the y x −4; 3 8. 1. From the equation y- (0, intercept is −3 . Thus, the −3) is on the 9. positive graph. We can obtain a second point by choosing a value for and finding the corresponding value x 10. True 11. False. If increases by 3, then increases by 2. x y for Let = 1 , then y. x (1) = 2 − 3 = −1 . Thus, y (1, −1) the point is also on the graph. Plotting the two points and connecting with a line yields 12. False. The is 8. The zero is −4 . y-intercept the graph below. ( = 2x + 3 f x ) a. Slope = 2; = 3 y-intercept b. Plot the point (0, 3). Use the slope to find 1) an additional point by moving 1 unit to the 13. right and 2 units up. 2. = − = 3 − 5 = −2 = 2 m y2 y1 − x2 x1 −1 − −3 3 2 c. average rate of change = 2 3. To find the let = 0: y-intercept, x 02 + 2 = 4 y d. increasing = 2 y ( ) = 5x − 4 g x So, the is (0, 2). y-intercept = 5; = −4 a. y-intercept To find the let = 0: x-intercept(s), y 14. b. Plot the (0, −4) . Use the slope to find + = 2(0) 4 x2 an additional point by moving 1 unit to the = 4 x2 = ±2 x So, the are (−2, 0) solution set is {50}. x-intercepts x = and (2, 0) . − = −15x + 4. 60x 900 2850 0 right and 5 units up. T 75x − 900 = 2850 h 75x = 3750 (−2) (−2)2 = − 4 = 4 − 4 = 5. f
Chapter 2 Linear and Quadratic Functions Section 2.1 7. y-intercept 1. From the equation y = 2 x − 3 , we see that the y- 8. −4; 3 intercept is −3 . Thus, the (0, −3) is on the graph. We can obtain a second point by choosing a value for x and finding the corresponding value 9. positive 10. True for y. Let x = 1 , then y = 2 (1) − 3 = −1 . Thus, 11. False. If x increases by 3, then y increases by 2. the point (1, −1) is also on the graph. Plotting the two points and connecting with a line yields the graph below. 12. False. The y-intercept is 8. The zero is −4 . 13. f ( x ) = 2x + 3 a. Slope = 2; y-intercept = 3 b. Plot the point (0, 3). Use the slope to find an additional point by moving 1 unit to the right and 2 units up. 2. m = y2 − y1 = 3 − 5 = −2 = 2 x2 − x1 −1 − 2 −3 3 3. To find the y-intercept, let x = 0: 02 + 2 y = 4 y = 2 So, the y-intercept is (0, 2). To find the x-intercept(s), let y = 0: x2 + 2(0) = 4 x2 = 4 x = ±2 14. c. average rate of change = 2 d. increasing g ( x ) = 5x − 4 a. = 5; y-intercept = −4 b. Plot the (0, −4) . Use the slope to find an additional point by moving 1 unit to the So, the x-intercepts are (−2, 0) 4. 60x − 900 = −15x + 2850 75x − 900 = 2850 75x = 3750 x = solution set is {50}. and (2, 0) . right and 5 units up. 5. f (−2) = (−2)2 − 4 = 4 − 4 = 0 h T 1)
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