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2.10 Warm Up Multiply the following polynomials. 1. (3p 4 – 5)(2p² + 4) 2. (8n² - 1)(3n² - 4n + 5) 3. (2s + 5)(s² + 3s – 1)
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2.10 Graphing y = ax 2 +c
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Vocabulary Quadratic Function: nonlinear, u-shaped graph (parabola) Standard Quadratic form: y = ax² + bx + c where a ≠ 0 Parent quadratic function: y = x² Vertex: highest/lowest point on graph Axis of symmetry: line that passes through vertex splitting into 2 equal parts (x = #)
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Recall: y = 3 x-4 +5 3 stretched it -4 moved it to the right +5 moved it up
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EXAMPLE 1 Graph y= ax 2 where a > 1 STEP 1 Graph y = 3x 2 First step is to find the vertex. 2 nd step is to pick one x-value on each side of the vertex – substitute these values in to find the y-coordinate. 3 rd step – plot the three points and draw your parabola. 4 th step – compare to parent function.
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EXAMPLE 2 Graph y = ax 2 where a < 1 Graph y = 1 4 – x2.x2. Compare the graph with the graph of y = x 2.
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EXAMPLE 3 Graph y = x 2 + c Graph y = x 2 + 5. Compare the graph with the graph of y = x 2. STEP 1
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GUIDED PRACTICE for Examples 1, 2 and 3 Graph the function. Compare the graph with the graph of x 2. 1. y= –4x 2 ANSWER 2. y = x 2 1 3 ANSWER
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GUIDED PRACTICE for Examples 1, 2 and 3 3. y = x 2 +2 Graph the function. Compare the graph with the graph of x 2. ANSWER
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EXAMPLE 4 Graph y = x 2 – 4. Compare the graph with the graph of y = x 2. 1 2 Graph y = ax 2 + c
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GUIDED PRACTICE for Example 4 Graph the function. Compare the graph with the graph of x 2. 4. y= 3x 2 – 6 5. y= –5x 2 + 1 6. y = x 2 – 2. 3 4
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GUIDED PRACTICE for Example 4 5. y= –5x 2 + 1
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GUIDED PRACTICE for Example 4 6. y = x 2 – 2. 3 4
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EXAMPLE 5 Standardized Test Practice How would the graph of the function y = x 2 + 6 be affected if the function were changed to y = x 2 + 2? A The graph would shift 2 units up. B The graph would shift 4 units up. C The graph would shift 4 units down. D The graph would shift 4 units to the left.
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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 = 0.09x 2 models the cross section of the reflective surface where x and y are measured in meters. Use the graph to find the domain and range of the function in this situation.
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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.
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GUIDED PRACTICE for Examples 5 and 6 Describe how the graph of the function y = x 2 +2 would be affected if the function were changed to y = x 2 – 2. 7. ANSWER The graph would be translated 4 units down. Domain: – 4 ≤ x ≤ 4, Range: 0 ≤y ≤1.44 ANSWER 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.
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