Warm Up

1. Preview Warm Up California Standards Lesson Presentation

2. x –2 –1 0 1 2 y 6 3 2 3 6 • Warm Up • 1. Evaluate x2 + 5x for x = 4 and x = –3. 36; –6 2. Generate ordered pairs for the function y = x2 + 2 for x-values –2, –1, 0, 1, and 2.

3. California Standards 21.0 Students graph quadratic functions and know that their roots are the x-intercepts.Also covered: 17.0

4. Vocabulary quadratic equation quadratic function parabola minimum value maximum value vertex

5. Solutions of the equation y = x2 are shown in the graph. Notice that the graph is not linear. The equation y = x2 is a quadratic equation. A quadratic equation in two variables can be written in the form y = ax2 + bx + c, where a, b, and c are real numbers and a ≠ 0. The equation y = x2 can be written as y = 1x2 + 0x + 0, where a = 1, b = 0, and c = 0.

6. Notice that the graph of y = x2 represents a function because each domain value is paired with exactly one range value. A function represented by a quadratic equation is a quadratic function.

7. For any quadratic equation in two variables • all points on its graph are solutions to the equation. • all solutions to the equation appear on its graph.

8. Substitute (4, 16) into 16 8+ 8 ? ? = = Since (4, 16) is a solution of , (4, 16)is on the graph. Additional Example 1A: Determining Whether a Point Is on a Graph Without graphing, tell whether each point is on the graph of (4, 16)  16 = 16

9. Substitute (–2, 10) into 10 2+ 8 ? ? = = Since (–2, 10) is a solution of , (–2, 10)is on the graph. Additional Example 1B: Determining Whether a Point Is on a Graph Without graphing, tell whether the function is on the graph of (–2, 10)  10 = 10

10. Substitute (–4, 0) into 0 8+ 8 ? ?  0 16 ≠ = = Since (–4, 0) is not a solution of , (–4, 0)is not on the graph. Additional Example 1C: Determining Whether a Point Is on a Graph Without graphing, tell whether each point is on the graph of (–4, 0)

11. Substitute (1, 1) into x2 + y = 2. 12 + 1 2 1 + 1 2 ? ? = = x2 + y = 2 Since (1, 1) is a solution of , (1, 1)is on the graph. Check It Out! Example 1a Without graphing, tell whether the point is on the graph of x2 + y = 2. (1, 1) x2 + y = 2  2 = 2

12. Substitute into x2 + y = 2. Since is not a solution of x2 + y = 2, is not on the graph. Check It Out! Example 1b Without graphing, tell whether the point is on the graph of x2 + y = 2. x2 + y = 2

13. (–3.5)2 + 10.5 2 12.25 + 10.5 2 ? ? = = Since (–3.5, 10.5) is not a solution of (–3.5, 10.5)is not on the graph. x2 + y =2, Check It Out! Example 1c Without graphing, tell whether the point is on the graph of x2 + y = 2. (–3.5, 10.5) x2 + y = 2 Substitute (–3.5, 10.5) into x2 + y = 2  22.75 ≠ 2

14. The graph of a quadratic function is a curve called a parabola. To graph a quadratic function, generate enough ordered pairs to see the shape of the parabola. Then connect the points with a smooth curve.

15. x y 4 3 –2 1 3 –1 0 0 1 3 1 4 3 2 Additional Example 2A: Graphing Quadratic Functions Graph the quadratic function. Make a table of values. Choose values of x and use them to find values of y. Graph the points. Then connect the points with a smooth curve.

16. x y –2 –16 –1 –4 0 0 1 –4 2 –16 Additional Example 2B: Graphing Quadratic Functions Graph the quadratic function. y = –4x2 Make a table of values. Choose values of x and use them to find values of y. Graph the points. Then connect the points with a smooth curve.

17. Helpful Hint When choosing values of x, be sure to choose positive values, negative values, and 0.

18. x y –2 6 –1 3 0 2 1 3 2 6 Check It Out! Example 2a Graph each quadratic function. y = x2 + 2 Make a table of values. Choose values of x and use them to find values of y. Graph the points. Then connect the points with a smooth curve.

19. x y –2 –11 –1 –2 0 1 1 –2 2 –11 Check It Out! Example 2b Graph the quadratic function. y = –3x2 + 1 Make a table of values. Choose values of x and use them to find values of y. Graph the points. Then connect the points with a smooth curve.

20. As shown in the graphs in Examples 2A and 2B, some parabolas open upward and some open downward. Notice that the only difference between the two equations is the value of a. When a quadratic function is written in the form y = ax2 + bx + c, the value of a determines the direction the parabola opens. • A parabola opens upward when a > 0. • A parabola opens downward when a < 0.

21. Write the function in the form y = ax2 + bx + c by solving for y. Add to both sides. Additional Example 3A: Identifying the Direction of a Parabola Tell whether the graph of the quadratic function opens upward or downward. Explain. Identify the value of a. Since a> 0, the parabola opens upward.

22. Additional Example 3B: Identifying the Direction of a Parabola Tell whether the graph of the quadratic function opens upward or downward. Explain. y = 5x– 3x2 Write the function in the form y = ax2 + bx + c. y = –3x2 + 5x a = –3 Identify the value of a. Since a< 0, the parabola opens downward.

23. Check It Out! Example 3a Tell whether the graph of each quadratic function opens upward or downward. Explain. f(x) = –4x2 – x + 1 f(x) = –4x2 – x + 1 Identify the value of a. a = –4 Since a < 0 the parabola opens downward.

24. + 5x2 + 5x2 Check It Out! Example 3b Tell whether the graph of each quadratic function opens upward or downward. Explain. y – 5x2= 2x – 6 Write the function in the form y = ax2 + bx + c by solving for y. Add 5x2 to both sides. y – 5x2= 2x – 6 y = 5x2 + 2x – 6 Identify the value of a. a = 5 Since a> 0 the parabola opensupward.

25. The minimum valueof a function is the least possible y-value for that function. The maximum valueof a function is the greatest possible y-value for that function. The highest or lowest point on a parabola is the vertex. Therefore, the minimum or maximum value of a quadratic function occurs at the vertex.

27. A. Additional Example 4: Identifying the Vertex and the Minimum or Maximum Identify the vertex of each parabola. Then give the minimum or maximum value of the function. B. The vertex is (–3, 2), and the minimum is 2. The vertex is (2, 5), and the maximum is 5.

28. a. b. Check It Out! Example 4 Identify the vertex of each parabola. Then give the minimum or maximum value of the function. The vertex is (–2, 5) and the maximum is 5. The vertex is (3, –1), and the minimum is –1.

29. Unless a specific domain is given, the domain of a quadratic function is all real numbers. One way to find the range of a quadratic function is by looking at its graph. For the graph of y = x2 – 4x + 5, the range begins at the minimum value of the function, where y = 1. All y-values greater than or equal to 1 appear somewhere on the graph. So the range is y 1.

30. Caution! You may not be able to see the entire graph, but that does not mean the graph stops. Remember that the arrows indicate that the graph continues.

31. Additional Example 5: Finding Domain and Range Find the domain and range. Step 1 The graph opens downward, so identify the maximum. The vertex is (–5, –3), so the maximum is –3. Step 2 Find the domain and range. D: all real numbers R: y ≤ –3

32. Check It Out! Example 5a Find the domain and range. Step 1 The graph opens upward, so identify the minimum. The vertex is (–2, –4), so the minimum is –4. Step 2 Find the domain and range. D: all real numbers R: y ≥ –4

33. Check It Out! Example 5b Find the domain and range. Step 1 The graph opens downward, so identify the maximum. The vertex is (2, 3), so the maximum is 3. Step 2 Find the domain and range. D: all real numbers R: y ≤ 3

34. Lesson Quiz: Part I 1. Without graphing, tell whether (3, 12) is on the graph of y = 2x2 – 5. 2. Graph y = 1.5x2. no

35. Lesson Quiz: Part II Use the graph for Problems 3-5. 3. Identify the vertex. 4. Does the function have a minimum or maximum? What is it? 5. Find the domain and range. (5, –4) maximum; –4 D: all real numbers; R: y ≤ –4