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# 10.2

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### 10.2

1. 1. y = ax + bx + c 2
2. 2. Properties of the Graph of aQuadratic FunctionThe graph of y = ax 2 + bx + c is a parabola that: Opens up if a > 0 and opens down if a < 0 Is narrower than the graph of y = x if a >1 and 2 wider if a <1 b Axis of Symmetry (AoS): x = - 2a b Has a vertex with an x-coordinate of - 2a Has a y-intercept of c. So, the point (0, c) is on the parabola.
3. 3. Example 1 Find the axis of symmetry and the vertexConsider the graph of the function y = – 2x2 + 12x – 7.a. Find the axis of symmetry.b. Find the vertex.SOLUTIONa. For the function y = – 2x2 + 12x – 7, a = – 2 and b = 12. b 12 Substitute –2 for a and 12 for b. x = – = – =3 Then simplify. 2a 2 ( – 2)
4. 4. Example 1 Find the axis of symmetry and the vertexANSWERThe axis of symmetry is the vertical line x = 3. bb. The x-coordinate of the vertex is – , or 3. 2aTo find the y-coordinate, substitute 3 for x in thefunction and simplify. 2y = – 2( 3) + 12( 3) – 7 = 11 Substitute 3 for x. Then simplify.ANSWER The vertex is ( 3, 11).
5. 5. Example 2 Graph y = ax2 + bx + cGraph y = 3x2 – 6x + 2.STEP 1 Determine whether the parabola opens up or down. Because a > 0, the parabola opens up.STEP 2 Find and draw the axis of symmetry: b –6 x = – = – =1 2a 2( 3)STEP 3 Find and plot the vertex. The x-coordinate of the vertex is – b , or 1. 2a
6. 6. Example 2 Graph y = ax2 + bx + c To find the y-coordinate, substitute 1 for x in the function and simplify. y = 3( 1)2 – 6 ( 1) + 2 = –1 So, the vertex is ( 1, – 1 ).STEP 4 Plot two points. Choose two x-values less than the x-coordinate of the vertex. Then find the corresponding y-values. x 0 –1 y 2 11
7. 7. Example 2 Graph y = ax2 + bx + cSTEP 5 Reflect the points plotted in Step 4 in the axis of symmetry.STEP 6 Draw a parabola through the plotted points.
8. 8. Minimum and Maximum ValuesFor y = ax 2 + bx + c , the y-coordinate of the vertex is theminimum value of the function if a > 0 or themaximum value of the function if a < 0.
9. 9. Example 3 Find the minimum or maximum valueTell whether the function y = – 3x2 – 12x + 10 has aminimum value or a maximum value. Then find theminimum or maximum value.SOLUTIONBecause a = – 3 and – 3 < 0, the parabola opens downand the function has a maximum value. To find themaximum value, find the vertex. b –12 bx = – = – = –2 The x-coordinate is – . 2a 2 ( – 3) 2a Substitute –2 for x. Theny = – 3(– 2) 2 – 12( – 2) + 10 = 22 simplify.
10. 10. Example 3 Find the minimum or maximum valueANSWERThe maximum value of the function is 22.
11. 11. Example 4 Find the minimum value of a functionSUSPENSION BRIDGESThe suspension cables between the two towers of theMackinac Bridge in Michigan form a parabola that canbe modeled by the graph of y = 0.000097x2 – 0.37x + 549where x and y are measured in feet. What is the heightof the cable above the water at its lowest point?SOLUTIONThe lowest point of the cable is at the vertex of theparabola. Find the x-coordinate of the vertex. Usea = 0.000097 and b = – 0.37.
12. 12. Example 4 Find the minimum value of a functionx = – b – 0.37 = – ≈ 1910 Use a calculator. 2a 2(0.000097)Substitute 1910 for x in the equation to find they-coordinate of the vertex.y = 0.000097 (1910)2 – 0.37 (1910) + 549 ≈ 196ANSWERThe cable is about 196 feet above the water at its lowestpoint.
13. 13. 10.2 Warm-UpTell whether the function has a minimum value or amaximum value. Then find he axis of symmetry and thevertex of the graph of the function. Finally graph thefunction.1. y = x 2 + 4x +82. y = -4x 2 + 4x + 8