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Unit 6 – Chapter 10

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Unit 6 – Chapter 10. Unit 6. Chapter 9 Review and Chap. 9 Skills Section 10.1 – Graph y = ax 2 + c Section 10.2 – Graph y = ax 2 + bx +c Section 10.3 – Solve Quadratic Eqns by Graphing Section 10.4 – Solve Quad. Eqns. w/Square roots Section 10.6 – Solve Quad. Eqns. With Quad. Formula

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Unit 6 – Chapter 10
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  1. Unit 6 – Chapter 10

  2. Unit 6 • Chapter 9 Review and Chap. 9 Skills • Section 10.1 – Graph y = ax2 + c • Section 10.2 – Graph y = ax2 + bx +c • Section 10.3 – Solve Quadratic Eqns by Graphing • Section 10.4 – Solve Quad. Eqns. w/Square roots • Section 10.6 – Solve Quad. Eqns. With Quad. Formula • Section 10.7-10.8 – Compare Linear, Exponential, and Quadratic Models

  3. Warm-Up – Chapter 10

  4. 1. The x-coordinate of a point where a graph crosses the x-axis is a (n) . ? ? x-intercept ANSWER = = 2. A(n)is a function of the formy = a bxwhere a 0,b > 0, and b 1. exponential function ANSWER Prerequisite Skills VOCABULARY CHECK Copy and complete the statement.

  5. 3. ANSWER Prerequisite Skills SKILLS CHECK Draw the blue figure. Then draw its image after a reflection in the red line.

  6. ANSWER 4. Prerequisite Skills SKILLS CHECK Draw the blue figure. Then draw its image after a reflection in the red line.

  7. Prerequisite Skills SKILLS CHECK Graph y1 = x and y2 = x – 5. Describe the similarities and differences in the graphs. ANSWER • Graphs are • Parallel • Have the same • Slope • Y2 is translated • Down 5 units

  8. ANSWER domain: all real numbers; range: all positive real numbers ANSWER Lesson 10.1, For use with pages 628-634 1.Graph the functiony = 2x. • Identify the domain and range of your graph in • Exercise 1.

  9. domain: all real numbers; range: all positive real numbers ANSWER Lesson 10.1, For use with pages 628-634 1.Graph the functiony = x2. ANSWER • Identify the domain and range of your graph in • Exercise 1.

  10. domain: all real numbers; range: all positive real numbers ANSWER Lesson 10.1, For use with pages 628-634 1.Graph the functiony = 3x2. ANSWER • Identify the domain and range of your graph in • Exercise 1.

  11. domain: all real numbers; range: all positive real numbers ANSWER Lesson 10.1, For use with pages 628-634 1.Graph the functiony = (0.1)x2. ANSWER • Identify the domain and range of your graph in • Exercise 1.

  12. domain: all real numbers; range: all positive real numbers > 2 ANSWER Lesson 10.1, For use with pages 628-634 1.Graph the functiony = x2 + 2 ANSWER • Identify the domain and range of your graph in • Exercise 1.

  13. domain: all real numbers; range: all negative numbers ANSWER Lesson 10.1, For use with pages 628-634 1.Graph the functiony = (-1)x2 ANSWER • Identify the domain and range of your graph in • Exercise 1.

  14. Vocabulary – 10.1 • Vertex of a Parabola • Lowest (or highest) point on a parabola • Axis of symmetry • Line that passes through the vertex • Divides the parabola into two symmetric parts • Quadratic Function • Non-linear function in the form y = ax2 + bx + c where a ≠ 0 • Parabola • U – shaped graph of a quadratic function • Parent quadratic function • Y = x2

  15. Notes – 10.1 – Graph y = ax2 + c • All Quadratic Function graphs look like what? • A parabola • What happens to the shape of the parabola if a >1? • It is a vertical stretch • It gets skinny! • What happens to the shape of a parabola if a < 1? • Vertical shrink • It gets fat (or “fluffy”!)! • What happens if a < 0? • The graph curves down. • What effect does “c” have on the graph of the function? • It translates it up or down.

  16. Examples 10.1

  17. Graph y= ax2 where a > 1 EXAMPLE 1 STEP 1 Make a table of values for y =3x2 STEP 2 Plot the points from the table.

  18. Graph y= ax2 where a > 1 EXAMPLE 1 STEP 3 Draw a smooth curve through the points. STEP 4 Compare the graphs of y = 3x2and y = x2. Both graphs open up and have the same vertex, (0, 0), and axis of symmetry,x = 0. The graph of y = 3x2 is narrower than the graph of y =x2 because the graph of y = 3x2 is a vertical stretch (by a factor of 3) of the graph of y =x2.

  19. Graph y =ax2 where a < 1 – x2. Graph y= Compare the graph with the graph of 1 1 y = x2. 4 4 – x2. Make a table of values for y = EXAMPLE 2 STEP 1

  20. Graph y =ax2 where a < 1 EXAMPLE 2 STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

  21. Graph y =ax2 where a < 1 1 1 1 4 4 4 andy =x2. Compare the graphs ofy = 1 – Both graphs have the same vertex (0, 0), and the same axis of symmetry,x = 0. However, the graph of x2. 4 – – x2. x2. is wider than the graph of y =x2 and it opens down. This is because the graph of is a vertical shrink with a refl ection in the x-axis of the graph of y =x2. by a factor of y = y = EXAMPLE 2 STEP 4

  22. EXAMPLE 3 Graph y = x2 + c Graph y = x2+ 5. Compare the graph with the graph of y = x2. STEP 1 Make a table of values for y =x2+ 5.

  23. EXAMPLE 3 Graph y = x2 + c STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

  24. EXAMPLE 3 Graph y = x2 + c STEP 4 Compare the graphs of y = x2+ 5and y = x2.Both graphs open up and have the same axis of symmetry, x = 0. However, the vertex of the graph of y =x2+ 5,(0, 5), is different than the vertex of the graph of y =x2,(0, 0), because the graph of y =x2 + 5 is a vertical translation (of 5 units up) of the graph of y =x2.

  25. for Examples 1, 2 and 3 GUIDED PRACTICE Graph the function. Compare thegraph with the graph ofx2. 1. y=–4x2 STEP 1 Make a table of values for y = –4x2

  26. for Examples 1, 2 and 3 GUIDED PRACTICE STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

  27. 2. y = x2 Make a table of values for y = x2 1 1 3 3 for Examples 1, 2 and 3 GUIDED PRACTICE STEP 1

  28. for Examples 1, 2 and 3 GUIDED PRACTICE STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

  29. for Examples 1, 2 and 3 GUIDED PRACTICE 3. y =x2 +2 STEP 1 Make a table of values for y = x2 +5

  30. for Examples 1, 2 and 3 GUIDED PRACTICE STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

  31. Graphy=x2–4.Compare the graph with the graph ofy=x2. Make a table of values fory=x2–4. 1 1 2 2 EXAMPLE 4 Graphy=ax2+c STEP1

  32. EXAMPLE 4 Graph y = ax2 + c STEP2 Plot the points from the table. STEP3 Draw a smooth curve through the points.

  33. Compare the graphs ofy=x2–4 andy=x2.Both graphs open up and have the same axis of symmetry, x = 0.However,the graph of y =x2– 4is wider and has a lower vertex than the graph of y =x2because the graph ofy =x2is a vertical shrink and a vertical translation of the graph of y=x2. 1 1 1 2 2 2 EXAMPLE 4 Graph y = ax2 + c STEP4

  34. for Example 4 GUIDED PRACTICE Graph the function. Compare thegraph with the graph ofx2. 4. y= 3x2 – 6 STEP1 Make a table of values fory=3x2 – 6.

  35. for Example 4 GUIDED PRACTICE STEP2 Plot the points from the table. STEP3 Draw a smooth curve through the points.

  36. for Example 4 GUIDED PRACTICE 5. y= –5x2 +1 STEP1 Make a table of values fory=–5x2 +1.

  37. for Example 4 GUIDED PRACTICE STEP2 Plot the points from the table. STEP3 Draw a smooth curve through the points.

  38. 6.y=x2–2. Make a table of values fory=x2–2. 3 3 4 4 for Example 4 GUIDED PRACTICE STEP1

  39. for Example 4 GUIDED PRACTICE STEP2 Plot the points from the table. STEP3 Draw a smooth curve through the points.

  40. The graph would shift 2 units up. A The graph would shift 4 units up. B The graph would shift 4 units down. C The graph would shift 4 units to the left. D EXAMPLE 5 Standardized Test Practice How would the graph of the function y = x2 + 6 be affected if the function were changed to y = x2 + 2?

  41. ANSWER The correct answer is C. A D B C EXAMPLE 5 Standardized Test Practice SOLUTION The vertex of the graph of y = x2 + 6 is 6 units above the origin, or (0, 6). The vertex of the graph of y = x2 + 2 is 2 units above the origin, or (0, 2). Moving the vertex from (0, 6) to (0, 2) translates the graph 4 units down.

  42. EXAMPLE 6 Use a graph SOLAR ENERGY A solar trough has a reflective parabolic surface that is used to collect solar energy. The sun’s rays are reflected from the surface toward a pipe that carries water. The heated water produces steam that is used to produce electricity. The graph of the function y = 0.09x2 models the cross section of the reflective surface where xandyare measured in meters. Use the graph to find the domain and range of the function in this situation.

  43. EXAMPLE 6 Use a graph SOLUTION STEP 1 Find the domain. In the graph, the reflective surface extends 5 meters on either side of the origin. So, the domain is 5 ≤ x ≤ 5. STEP 2 Find the range using the fact that the lowest point on the reflective surface is (0, 0) and the highest point, 5, occurs at each end. y = 0.09(5)2= 2.25 Substitute 5 for x. Then simplify. The range is 0≤ y ≤ 2.25.

  44. Describe how the graph of the function y = x2+2 would be affected if the function were changed to y=x2– 2. 7. ANSWER The graph would be translated 4 unit down. for Examples 5 and 6 GUIDED PRACTICE SOLUTION The vertex of the graph of y = x2 + 2 is 6 units above the origin, or (0, 2). The vertex of the graph of y = x2 – 2 is 2 units down above the origin, or (0, – 2). Moving the vertex from (0, 2) to (0, – 2) translates the graph 4 units down.

  45. What if In example 6 ,suppose the reflective surface extends just 4 meters on either side of the origin.Find the range of the function in this situation. 8. for Examples 5 and 6 GUIDED PRACTICE SOLUTION STEP 1 Find the domain. In the graph, the reflective surface extends 4 meters on either side of the origin. So, the domain is – 4 ≤ x ≤ 4.

  46. for Examples 5 and 6 GUIDED PRACTICE STEP 2 Find the range using the fact that the lowest point on the reflective surface is (0, 0) and the highest point 4, occurs at each end. y = 0.09(4)2= 144 Substitute 5 for x. Then simplify. The range is – 4 ≤ x ≤ 4, 0 ≤y ≤1.44.

  47. Warm-Up – 10.2

  48. 7 ANSWER 17 ANSWER Lesson 10.2, For use with pages 635-640 Evaluate the expression. 1.x2– 2 whenx = 3 2. 2x2 + 9 whenx = 2

  49. 3. Martin is replacing a square patch of counter top. The area of the patch is represented by A = s2. What is the area of the patch if the side length is 2.5 inches? 6.25 in.2 ANSWER Lesson 10.2, For use with pages 635-640 Evaluate the expression.

  50. Vocabulary –10.2 • Maximum Value • Highest possible value of a graph • Not all graphs have a maximum value • Minimum Value • Smallest possible value of a graph • Not all graphs have a minimum value • Vertex • AKA the “maximum” or “minimum” value of a parabola