Warm Up

1. Graph y = ax2 + c Warm Up Lesson Presentation Lesson Quiz

2. ANSWER domain: all real numbers; range: all positive real numbers ANSWER Warm-Up 1.Graph the functiony = 2x. • Identify the domain and range of your graph in • Exercise 1.

3. Example 1 Graph y = 3x2. Compare the graph with the graph of y = x2. STEP 1 Make a table of values for y =3x2. STEP 2 Plot the points from the table.

4. 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.

5. – 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 STEP 2 Plot the points from the table.

6. 1 1 1 4 4 4 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 Compare the graphs ofy = is wider than the graph of y =x2 and it opens down. This is because the graph of andy =x2. is a vertical shrink with a reflection in the x-axis of the graph of y =x2. by a factor of y = y = Example 2 STEP 3 Draw a smooth curve through the points. STEP 4

7. Example 3 Graph y = x2+ 5. Compare the graph with the graph of y = x2. STEP 1 Make a table of values for y =x2+ 5. STEP 2 Plot the points from the table.

8. Example 3 STEP 3 Draw a smooth curve through the points. 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.

9. ANSWER Guided Practice Graph the function. Compare thegraph with the graph ofx2. 1. y=–4x2

10. 2. y = x2 ANSWER 1 3 Guided Practice Graph the function. Compare thegraph with the graph ofx2.

11. ANSWER Guided Practice Graph the function. Compare thegraph with the graph ofx2. 3. y =x2 +2

12. 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 STEP1 STEP2 Plot the points from the table.

13. 1 1 1 Compare the graphs ofy=x2–4andy=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=x2 – 4is a vertical shrink and a vertical translation of the graph of y=x2. 2 2 2 Example 4 STEP3 Draw a smooth curve through the points. STEP4

14. The graph is a vertical stretch (by a factor of 3) with a vertical translation (of 6 units down) of the graph ofy = x2. Guided Practice Graph the function. Compare thegraph with the graph ofx2. 4. y=3x2 – 6

15. The graph is a vertical stretch (by a factor of 5) with a vertical translation (of 1 unit up) and a reflection in thex-axis of the graph ofy = x2. Guided Practice 5. y=–5x2 +1

16. 6.y=x2–2. The graph is a vertical shrink (by a factor of ) with a vertical translation (of 2 units down) of the graph of y = x2. 3 3 4 4 Guided Practice

17. 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 How would the graph of the function y = x2 + 6 be affected if the function were changed to y = x2 + 2?

18. ANSWER The correct answer is C. A D B C Example 5 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.

19. 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 units down. Guided Practice

20. Example 6 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 = 0.09x2models the cross section of the reflective surface where xandyaremeasured in meters. Use the graph to find the domain and range of the function in this situation.

21. Example 6 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 occurs at x=5 or x=5 . y = 0.09(5)2= 2.25 Substitute 5 for x. Then simplify. The range is 0≤ y ≤ 2.25.

22. WHAT IF? In Example 6, suppose the reflective surface extends just 4 meters on either side of the origin. Find the domain and range of the function in this situation. 8. ANSWER Domain: – 4 ≤ x ≤ 4, Range: 0 ≤y ≤1.44 Guided Practice

23. ANSWER ANSWER It would be shifted down 6 units. Lesson Quiz 1. Graphy = –0.5x2 + 2. 2. How would the graph of the functiony = –2x2 + 3 beaffected if the function were changedtoy = –2x2 – 3?

24. ANSWER about 1.8 sec Lesson Quiz 3. A pinecone falls about 50 feet from the branch of a pine tree. Its height (in feet) can be modeled by the function h(t) = 16t2 + 50, where t is the time in seconds. How long does it take to land on the ground?