1. Quadratic Functions PreCalculus 3-3

2. The graph of any quadratic function is called a parabola. Parabolas are shaped like cups, as shown in the graph below. If the coefficient of x 2 is positive, the parabola opens upward; otherwise, the parabola opens downward. The vertex (or turning point) is the minimum or maximum point. Graphs of Quadratic Functions

3. Form of a Quadratic Function

4. Graphing Parabolas With Equations in Standard Form To graph f (x)  a(x  h) 2  k: 1.Determine whether the parabola opens upward or downward. If a  0, it opens upward. If a  0, it opens downward. 2.Determine the vertex of the parabola. The vertex is (h, k). 3.Find any x-intercepts by replacing f (x) with 0. Solve the resulting quadratic equation for x. 4.Find the y-intercept by replacing x with zero. 5.Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a cup.

5. Step 1 Determine how the parabola opens. Note that a, the coefficient of x 2, is -2. Thus, a  0; this negative value tells us that the parabola opens downward. Standard form f (x)  a(x  h) 2  k a  -2 h  3 k  8 Given equation f (x)   2(x  3) 2  8 Solution We can graph this function by following the steps in the preceding box. We begin by identifying values for a, h, and k. Example Graph the quadratic function f (x)   2(x  3) 2  8.

6.   2(x  3) 2  8 Find x-intercepts, setting f (x) equal to zero. Step 3 Find the x-intercepts. Replace f (x) with 0 in f (x)   2(x  3) 2  8. Step 2 Find the vertex. The vertex of the parabola is at (h, k). Because h  3 and k  8, the parabola has its vertex at (3, 8). (x  3) 2  4 Divide both sides by 2. (x  3)   2 Apply the square root method. x  3  2 or x  3  2 Express as two separate equations. x  1 or x  5 Add 3 to both sides in each equation. The x-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0). 2(x  3) 2  8 Solve for x. Add 2(x  3) 2 to both sides of the equation. Example

7. Step 5 Graph the parabola. With a vertex at (3, 8), x-intercepts at 1 and 5, and a y-intercept at –10, the axis of symmetry is the vertical line whose equation is x  3.  f   2(0  3) 2  8   2(  3) 2  8   2(9)  8   10 Step 4 Find the y-intercept. Replace x with 0 in f (x)   2(x  3) 2  8. The y-intercept is –10. The parabola passes through (0,  10). Example

8. The Vertex of a Parabola Whose Equation Is f (x)  ax 2  bx  c Consider the parabola defined by the quadratic function f (x)  ax 2  bx  c. The parabola's vertex is at

9. Example Solution: Step 1 Determine how the parabola opens. Note that a, the coefficient of x 2, is -1. Thus, a  0; this negative value tells us that the parabola opens downward. Step 2 Find the vertex. We know the x-coordinate of the vertex is x = - b/(2a). We identify a, b, and c to substitute the values into the equation for the x-coordinate: x = -b/(2a) = -6/2(-1)=3. The x-coordinate of the vertex is 3. We substitute 3 for x in the equation of the function to find the y-coordinate: y=f(3) = -(3)^2+6(3)-2=-9+18-2=7, the parabola has its vertex at (3,7). Graph the quadratic function f (x)   x 2  6x 

10. Step 3 Find the x-intercepts. Replace f (x) with 0 in f (x)   x 2  6x  2. 0 =  x 2  6x  2 Graph the quadratic function f (x)   x 2  6x  Example

11. Step 4 Find the y-intercept. Replace x with 0 in f (x)   x 2  6x  2. f   0 2  6 0  2    The y-intercept is –2. The parabola passes through (0,  2).  Step 5 Graph the parabola. Graph the quadratic function f (x)   x 2  6x 

12. Minimum and Maximum: Quadratic Functions Consider f(x) = ax 2 + bx +c. 1.If a > 0, then f has a minimum that occurs at x = -b/(2a). This minimum value is f(-b/(2a)). 2.If a < 0, the f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a)).

13. Maximizing or Minimizing Quadratic Functions 1.Read the problem carefully and decide which quantity is to be maximized or minimized. 2.Use the conditions of the problem to express the quantity as a function in one variable. 3.Rewrite the function in the form f(x) = ax 2 + bx +c. 4.Calculate -b/(2a). If a > 0, then f has a minimum that occurs at x = -b/(2a). This minimum value is f(-b/(2a)). If a < 0, the f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a)). 5.Answer the question posed in the problem.

14. Quadratic Functions

16. Form of a Quadratic Function

17. Homework Pages 171 – 172 10 – 18even, 36, 39, 52, 54