1. 1.4 More Quadratic Functions and Applications 1 In the previous section, we transformed a quadratic function from the form f(x)=ax 2 +bx+c to the form f(x)=a(x–h) 2 +k in order to more easily identify the vertex, (h,k). We did this by the process known as completing the square. In this section we will show you another way to perform this transformation. Using the process of completing the square, we can obtain the vertex with the following procedure: 1.Identify the real number values a, b, and c. If a is positive, the parabola will open upward. If a is negative, the parabola will open downward. Next Slide

2. 1.4 More Quadratic Functions and Applications 2 1.Identify a, b, and c. Solution: a=2, b=-8, and c=5 Answer: The vertex is (2,-3). Your Turn Problem #1

3. 1.4 More Quadratic Functions and Applications 3 Example 2. Graph f(x)= –3x 2 +6x+2. Solution: The vertex is (1,5). 1. Find the vertex. 2.Make a table. Use two numbers to the right of 1 and two numbers to the left of 1. x y 2 2 3 -7 0 2 -1 -7

4. 1.4 More Quadratic Functions and Applications 4 Your Turn Problem #2 Vertex (3,-4) x y 4 -2 5 4 2 -2 1 4

5. 1.4 More Quadratic Functions and Applications 5 A parabola can have two x intercepts (figure 1), one x intercept (figure 2), or no x intercepts (figure 3). For those that do, we usually want to know what they are. To find x intercepts: set the function equal to 0, and solve the resulting equation for x. Note: If the values of x found by setting the equation equal to 0 are not real, then the parabola has no x intercepts. x Intercepts of a Parabola x f(x) x Figure 1Figure 2 Figure 3 x f(x) Next Slide

6. 1.4 More Quadratic Functions and Applications 6 Example 3.Find the x intercepts and the vertex of f(x)= –x 2 +6x–5. Solution: To find the x intercepts, set the function equal to zero and solve. The x-intercepts are (5,0) and (1,0). –x 2 + 6x – 5 = 0 (x – 5)(x – 1) = 0 x – 5=0 x=5 x – 1=0 x=1 x 2 –6x +5 = 0 To find the vertex, find x = -b/2a. a= –1, b=6, and c= –5 Find the x-intercepts and the vertex of f(x)=2x 2 – 36x+160. Your Turn Problem #3 Answer: The x-intercepts are (8,0) and (10,0).

7. 1.4 More Quadratic Functions and Applications 7 There is a correspondence between x-intercepts and zeros, which are x-values that have corresponding y-values that equal 0. 2 and –3 are zeros of the function f(x)=x 2 +x–6. The graph of this quadratic function is a parabola with x-intercepts of (2,0) and (-3,0). The procedure for finding zeros of a function is the same as that of finding x-intercepts for the graph of a function, except the answer is not an ordered pair, but instead, merely the x-value. Example 4.Find the zeros of f(x)= –x 2 +6x+6. Solution: To find the zeros, set the function equal to zero and solve. This trinomial does not factor. Use quadratic formula or completing the square to solve. –x 2 +6x–6=0 x 2 – 6x– 6 = 0 The zeros of the function are: and Find the zeros of f(x)=12x 2 – x – 20. Your Turn Problem #4

8. 1.4 More Quadratic Functions and Applications 8 Applications of Quadratic Functions When quadratic functions are used to model real life situations, we see that the model will either have a lowest or highest value at the vertex. 1.The x-value of the vertex indicates where the minimum or maximum occurs. 2.F(x) gives the minimum or maximum value of the function. x f(x) x minimum maximum Next Slide

9. 1.4 More Quadratic Functions and Applications 9 Example 5. Dog Homes R Us earns a yearly profit according to the function f(x)= –0.4x 2 +80x–200, where x is the number of dog houses produced and sold and f(x) is the profit in thousands of dollars. Find the maximum profit the company can earn in a week. Solution: Since this is a quadratic function, in which the value of ‘a’ is negative, we know that the parabola of the function’s graph opens downward. We are therefore guaranteed a maximum value, which would correspond to the function’s maximum profit. Understand that the x-value of the vertex would give the number of dog houses produced and sold that would yield the maximum profit. It is the y-value of the vertex that is the maximum profit. (Recall this value is in thousands of dollars.) a= –0.4, b=80, c= –200 Remember, this maximum value is in thousands of dollars, so 3800 thousand dollars is more commonly known as $3,800,000. Layton Ltd. earns a weekly profit according to the function f(x)= –0.001x 2 +2.45x–525, where x is the number of salt shakers produced and sold and f(x) is the profit in thousands of dollars. Find the maximum profit the corporation can earn in a year. Your Turn Problem #5 Answer: The maximum profit is $975,625. The End B.R. 1-5-07