1. Introduction Quadratic functions are used to model various situations. Some situations are literal, such as determining the shape of a parabola, and some situations involve applying the key features of quadratics to real-life situations. For example, an investor might want to predict the behavior of a particular mutual fund over time, or an NFL scout might want to determine the maximum height of a ball kicked by a potential football punter. In this lesson, we will look specifically at the vertex form of a quadratic, f(x) = a(x – h) 2 + k, where the vertex is the point (h, k). The vertex can be read directly from the equation. 1 5.3.3: Creating and Graphing Equations Using Vertex Form

2. Key Concepts Standard form, intercept form, and vertex form are equivalent expressions written in different forms. Standard form: f(x) = ax 2 + bx + c, where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant term Intercept form: f(x) = a(x – p)(x – q), where p and q are the zeros of the function Vertex form: f(x) = a(x – h) 2 + k, where the vertex of the parabola is the point (h, k) 2 5.3.3: Creating and Graphing Equations Using Vertex Form

3. Key Concepts, continued To identify the vertex directly from an equation in vertex form, identify h (the x-coordinate of the vertex) and k (the y-coordinate of the vertex). Note that the original equation in vertex form has the quantity x – h, so if the equation has a subtraction sign then the value of h is h. This is true because x – (–h) simplifies to x + h. 3 5.3.3: Creating and Graphing Equations Using Vertex Form

4. Key Concepts, continued However, if the quantity is written as x + h, the value of h is –h. A quadratic function in standard form can be created from vertex form, f(x) = a(x – h) 2 + k, where (h, k) is the vertex of the quadratic. To do so, distribute and simplify by combining like terms. For example, f(x) = 3(x – 2) 2 + 4 becomes f(x) = 3x 2 – 12x + 16. 4 5.3.3: Creating and Graphing Equations Using Vertex Form

5. Key Concepts, continued A quadratic function in vertex form can be created from standard form, f(x) = ax 2 + bx + c. To do so, complete the square, or determine the value of c that would make ax 2 + bx + c a perfect square trinomial. To complete the square, take the coefficient of the linear term, divide by the product of 2 and the coefficient of the quadratic term, and square the quotient. 5 5.3.3: Creating and Graphing Equations Using Vertex Form

6. Key Concepts, continued 6 5.3.3: Creating and Graphing Equations Using Vertex Form

7. Key Concepts, continued Since the quotient of b and 2a is a constant term, we can combine it with the constant c to get the equation, where For example, f(x) = 2x 2 – 12x + 22 becomes f(x) = 2(x – 3) 2 + 4. 7 5.3.3: Creating and Graphing Equations Using Vertex Form

8. Key Concepts, continued When graphing a quadratic using vertex form, if the vertex is the y-intercept, choose two pairs of symmetric points to plot in order to sketch the most accurate graph. 8 5.3.3: Creating and Graphing Equations Using Vertex Form

9. Common Errors/Misconceptions forgetting to make sure the coefficient of the quadratic term, x 2, is 1 before completing the square 9 5.3.3: Creating and Graphing Equations Using Vertex Form

10. Guided Practice Example 2 Determine the equation of a quadratic function that has a minimum at (–4, –8) and passes through the point (–2, –5). 10 5.3.3: Creating and Graphing Equations Using Vertex Form

11. Guided Practice: Example 2, continued 1.Substitute the vertex into f(x) = a(x – h) 2 + k. 11 5.3.3: Creating and Graphing Equations Using Vertex Form f(x) = a(x – h) 2 + kVertex form f(x) = a[x – (–4)] 2 + (–8) Substitute (–4, –8) for h and k. f(x) = a(x + 4) 2 – 8Simplify.

12. Guided Practice: Example 2, continued 2.Substitute the point (–2, –5) into the equation from step 1 and solve for a. 12 5.3.3: Creating and Graphing Equations Using Vertex Form f(x) = a(x + 4) 2 – 8Equation –5 = a[(–2) + 4] 2 – 8 Substitute (–2, –5) for x and f(x). –5 = a(2) 2 – 8Simplify. –5 = 4a – 8 3 = 4a

13. Guided Practice: Example 2, continued 3.Substitute a into the equation from step 1. f(x) = a(x + 4) 2 – 8 The equation of the quadratic function with a minimum at (–4, –8) and passing through the point (–2, –5) is 13 5.3.3: Creating and Graphing Equations Using Vertex Form ✔

14. Guided Practice: Example 2, continued 14 5.3.3: Creating and Graphing Equations Using Vertex Form

15. Guided Practice Example 4 Sketch a graph of the quadratic function y = (x + 3) 2 – 8. Label the vertex, the axis of symmetry, the y-intercept, and one pair of symmetric points. 15 5.3.3: Creating and Graphing Equations Using Vertex Form

16. Guided Practice: Example 4, continued 1.Identify the vertex and the equation of the axis of symmetry. Given the vertex form of a quadratic function, f(x) = a(x – h) 2 + k, the vertex is the point (h, k). The vertex of the quadratic y = (x + 3) 2 – 8 is (–3, –8). The axis of symmetry extends through the vertex. The equation of the axis of symmetry is x = –3. 16 5.3.3: Creating and Graphing Equations Using Vertex Form

17. Guided Practice: Example 4, continued 2.Find the y-intercept. The parabola crosses the y-axis when x = 0. Substitute 0 for x to find y. The y-intercept is the point (0, 1). 17 5.3.3: Creating and Graphing Equations Using Vertex Form y = (x + 3) 2 – 8Original equation y = (0 + 3) 2 – 8Substitute 0 for x. y = 3 2 – 8Simplify. y = 1

18. Guided Practice: Example 4, continued 3.Find an extra point to the left or right of the axis of symmetry. Choose an x-value and substitute it into the equation to find the corresponding y-value. Typically, choosing x = 1 or x = –1 is simplest arithmetically, if these numbers aren’t already a part of the vertex or axis of symmetry. In this case, let’s use x = 1. 18 5.3.3: Creating and Graphing Equations Using Vertex Form

19. Guided Practice: Example 4, continued The parabola passes through the point (1, 8). x = 1 is 4 units to the right of the axis of symmetry, x = –3. 4 units to the left of the axis of symmetry and horizontal to (1, 8) is the symmetric point (–7, 8). 19 5.3.3: Creating and Graphing Equations Using Vertex Form y = (x + 3) 2 – 8Original equation y = (1 + 3) 2 – 8Substitute 1 for x. y = 4 2 – 8Simplify. y = 8

20. Guided Practice: Example 4, continued 4.Plot the points you found in steps 2 and 3 and their symmetric points over the axis of symmetry. 20 5.3.3: Creating and Graphing Equations Using Vertex Form

21. Guided Practice: Example 4, continued 21 5.3.3: Creating and Graphing Equations Using Vertex Form ✔

22. Guided Practice: Example 4, continued 22 5.3.3: Creating and Graphing Equations Using Vertex Form