The document discusses modeling quadratic functions based on given information about the graph. It provides an example of a parabola with the vertex at (2, -3) and a point (4, 1) given. Using the point-slope form, the quadratic function for this parabola is determined to be y = 1(x - 2)(x - 4) based on the given vertex and point. It also shows writing the equation in intercept form using the x-intercepts -2 and 3 and a point (-1,2).

1. 5.8 Modeling with Quadratic Functions

2. In Lesson 5.1 you learned how to graph a given quadratic function. In this lesson you will write quadratic functions when given information about their graphs.

3. For example, here’s a parabola. Notice the coordinates of the vertex and a point on the graph are given. Vertex (2, -3) (4, 1) Start with Now use the other given point (4, 1) as replacements for (x,y) a = 1 Thus a quadratic function for the parabola is

4. Write the equation of this parabola. Notice the x-intercepts of the parabola. -2 3 Start with the intercept form: y = a(x – p)(x – q) y = a(x – -2 )(x – 3 ) (-1,2) Now use the coordinates of a point (-1, 2) and substitute them for (x,y) and solve for a . 2 = a(- 1 + 2 )(- 1 – 3 ) a = -1/2