## Presentation on theme: "Chapter 9 Section 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley."— Presentation transcript:

More on Graphing Quadratic Equations: Quadratic Functions Graph quadratic equations of the form y = ax 2 + bx + c ( a  0). Use a graph to determine the number of real solutions of a quadratic equation. Use a quadratic function to solve an application. 1 1 3 3 2 29.59.5

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley More on Graphing Quadratic Equations; Quadratic Functions In Section 5.4, we graphed the quadratic equation y = x 2. By plotting points, we obtained the graph of a parabola shown here. Slide 9.5 - 3 Recall the lowest (or highest point if the parabola opens downward) point on the graph is called the vertex of the parabola. The vertical line through the vertex is called the axis, or axis of symmetry. The two halves of the parabola are mirror images of each other across this axis.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1 Objective 1 Graph quadratic equations of the form y = ax 2 + bx + c (a  0). Slide 9.5 - 4

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph quadratic equations of the form y = ax 2 + bx + c (a  0). Every equation of the form y = ax 2 + bx + c with a  0, has a graph that is a parabola. The vertex is the most important point to locate when graphing a quadratic equation. Slide 9.5 - 5 Earlier in the course we solved linear equations in one variable that were of the form Ax + B = C ; then graphed linear equations in two variables that were of the form y = mx + b. Now, we are ready to do the same sort of thing for quadratic equations. We know how to solve ax 2 + bx + c = 0; and now we graph y = ax 2 + bx + c.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph quadratic equations of the form y = ax 2 + bx + c (a  0). (cont’d) Slide 9.5 - 6 A procedure for graphing quadratic equations follows: Step 1: Find the vertex. Let x =, and find the corresponding y-value by substituting for x in the equation. Step 2: Find the y-intercept. Step 3: Find the x-intercepts (if they exist). Step 4: Plot the intercepts and the vertex. Step 5: Find and plot additional ordered pairs near the vertex and intercepts as needed, using symmetry about the axis of the parabola.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Graph y = x 2 + 2x – 8. Solution: Graphing a Parabola by Finding the Vertex and Intercepts Slide 9.5 - 7

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph y = –x 2 + 2x + 4. EXAMPLE 2 Graphing a Parabola Slide 9.5 - 8 Solution:

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2 Objective 2 Use a graph to determine the number of real solutions of a quadratic equation. Slide 9.5 - 9

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Use a graph to determine the number of real solutions of a quadratic equation. Using the vertical line test from Section 3.6, we see that the graph of an equation of the form y = ax 2 + bx + c is the graph of a function. A function defined by an equation of the form  (x) = ax 2 + bx + c (a  0) is called a quadratic function. The domain (possible x-values) of a quadratic function is the set of all real numbers, or (– ,  ); the range (the resulting y-values) can be determined after the function is graphed. Slide 9.5 - 10

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Solution: There are two real solutions, {±2}, that correspond to the graph. Determining the Number of Real Solutions from Graphs Slide 9.5 - 11 Decide from the graph how many real number solutions the corresponding equation x 2 − 4 = 0 has. Give the solution set.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3 Objective 3 Slide 9.5 - 12 Use a quadratic function to solve an application.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Solution: Let y = 48 ft, and x = 175 ft, solve for a. Finding the Equation of a Parabolic Satellite Dish Slide 9.5 - 13 Suppose that a radio telescope has a parabolic dish shape with diameter 350 ft and depth 48 ft. Find the equation of the graph of a cross section.