1. Section 3.1 Quadratic Functions; Parabolas Copyright ©2013 Pearson Education, Inc.

2. Objectives  Determine if a function is quadratic  Determine if the graph of a quadratic function is a parabola that opens down  Determine if the vertex of the graph of a quadratic function is a maximum or a minimum  Determine if a function increases or decreases over a given interval  Find the vertex of the graph of a quadratic function  Graph a quadratic function  Write the equation of a quadratic function given information about its graph  Find the vertex form of the equation of a quadratic function

3. Graph of a Quadratic Function The general form of a quadratic function is f(x) = ax 2 + bx + c where a, b, and c are real numbers with a ≠ 0. The graph of the quadratic function f(x) = ax 2 + bx + c is a parabola with a “turning point” called the vertex. The parabola opens upward (is concave up) if a is positive and the vertex is a minimum point. The parabola opens downward (is concave down) if a is negative and the vertex is a maximum point.

4. Increasing and Decreasing Functions A function f is increasing on an interval if, for any x 1 and x 2 in the interval, when x 2 > x 1, it is true that f(x 2 ) > f(x 1 ). A function f is decreasing on an interval if, for any x 1 and x 2 in the interval, when x 2 > x 1, it is true that f(x 2 ) < f (x 1 ).

6. Example Find the vertex and graph the quadratic function f(x) = –2x 2 – 4x + 6. Solution Since the coefficient of x 2 is –2 the parabola opens downward. The x-coordinate of the vertex is And the y-coordinate of the vertex is Thus the vertex is (–1, 8). f(–1) = –2(–1) 2 – 4(–1) + 6 = 8

7. Example (cont) Find the vertex and graph the quadratic function f(x) = –2x 2 – 4x + 6. Solution The x-intercepts can be found by setting f(x) = 0 and solving for x:

8. Example (cont) Solution The y-intercepts can be found by computing f(0). The axis of symmetry is the vertical line x = –1.

9. Example—Spreadsheet solution The graph of the function f(x) = –2x 2 – 4x + 6 using Excel.

10. Example A ball is thrown upward at 64 feet per second from the top of an 60-foot-high building. a. Write the quadratic function that models the height (in feet) of the ball as a function of the time t (in seconds). Solution a. The model has the form S(t) = –16t 2 + v 0 t + h 0, where v 0 = 64 and h 0 = 60. Thus the model is S(t) = –16t 2 + 64t + 60 (feet)

11. Example (cont) b. Find the t-coordinate and S-coordinate of the vertex of the graph of this quadratic function. Solution b. The height S is the function of time t, and the t-coordinate of the vertex is The S coordinate of the vertex is the value of S at t = 2, so S = –16(–2) 2 + 64(2) + 60 = 124. The vertex is (2, 124).

12. Example (cont) c. Graph the model. Solution S = –16t 2 + 64t + 60

13. Example (cont) d. Explain the meaning of the coordinates of the vertex for this model. Solution d. The graph is a parabola that opens down, so the vertex is the highest point on the graph and the function has a minimum there. The t-coordinate of the vertex, 2, is the time (in seconds) at which the ball reaches the maximum height, and the S-coordinate, 124, is the maximum height (in feet) that the ball reaches.

14. Graph of a Quadratic Function In general, the graph of the function y = a(x  h) 2 + k is a parabola with its vertex at the point (h, k). The parabola opens upward if a > 0, and the vertex is a minimum. The parabola opens downward if a < 0, and the vertex is a maximum. The axis of symmetry of the parabola has equation x = h. The a is the same as the leading coefficient in y = ax 2 + bx + c, so the larger the value of |a|, the narrower the parabola will be.

15. Example Right Sports Management had its monthly maximum profit, $450,000, when it produced and sold 5500 Waist Trimmers. Its fixed cost is $155,000. If the profit can be modeled by a quadratic function of x, the number of Waist Trimmers produced and sold each month, find this quadratic function P(x). Solution When 0 units are produced, the cost is $155,000 and the revenue is $0. Thus, the profit is –$155,000 when 0 units are produced, and the y-intercept of the graph of the function is (0, –155,000).

16. Example (cont) Right Sports Management had its monthly maximum profit, $450,000, when it produced and sold 5500 Waist Trimmers. Its fixed cost is $155,000. If the profit can be modeled by a quadratic function of x, the number of Waist Trimmers produced and sold each month, find this quadratic function P(x). Solution The vertex of the graph of the quadratic function is (5500, 450,000). Using these points gives P(x) = a(x  5500) 2 + 450,000 and –155,000 = a(0 – 5500) 2 + 450,000 which gives a = –0.02 Thus, the quadratic function that models the profit is P (x) = –0.02(x – 5500) 2 + 450,000, or P (x) = –0.02x 2 + 220x – 155,000, where P(x) is in dollars and x is the number of units produced and sold.

17. Example If the points in the table lie on a parabola, write the equations whose graph is the parabola. Solution The vertex of the parabola is (3,  5). The equation is y = (x – 3) 2 – 5. x 11 012 y114 11 44

18. Example Write the vertex form of the equation of the quadratic function from the general form y = 2x 2 – 8x + 3. Solution The vertex is at and y = 2(2 2 ) – 8(2) + 3 = –5. Thus, the vertex form of the equation is y = 2(x – 2) 2 – 5.