Explanation Result used: If the general quadratic equation is a x 2 + b x + c , then the x intercept is x = − b 2 a . Calculation: The given equation f ( x ) = − 2 3 ( x + 1 2 ) 2 + 10 can be written as follows, f ( x ) = − 2 3 ( x 2 + 2 ⋅ 1 2 x + 1 4 ) + 10 = ( − 2 3 ⋅ x 2 ) + ( − 2 3 ⋅ x ) + ( − 2 3 ⋅ 1 4 ) + 10 = − 2 3 x 2 − 2 3 x − 2 12 + 10 On further simplification, f ( x ) = − 2 3 x 2 − 2 3 x − 2 12 + 120 12 = − 2 3 x 2 − 2 3 x + 118 12 = − 2 3 x 2 − 2 3 x + 59 6 Compare the given equation f ( x ) = − 2 3 x 2 − 2 3 x + 59 6 into the standard quadratic equation then a = 1 2 , b = 3 and c = 29 2 . Substitute a = − 2 3 and b = − 2 3 in x = − b 2 a and obtain the x coordinate of the vertex. x = − ( − 2 3 ) 2 ( − 2 3 ) = − 1 2 That is, the x intercept is x = − 1 2 . Substitute x = − 1 2 in f ( x ) = − 2 3 ( x + 1 2 ) 2 + 10 to compute the y coordinate of the vertex. f ( − 1 2 ) = − 2 3 ( − 1 2 + 1 2 ) 2 + 10 = − 2 3 ( 0 ) 2 + 10 = 0 + 10 = 10 That is, the vertex of the parabola is ( − 1 2 , 10 ) . Obtain the x and y intercept of the parabola. Substitute x = 0 in f ( x ) = − 2 3 x 2 − 2 3 x + 59 6 to compute the y intercept. f ( 0 ) = − 2 3 ( 0 ) 2 − 2 3 ( 0 ) + 59 6 = 0 − 0 + 59 6 = 59 6 Therefore, the y intercept is ( 0 , 59 6 ) . Substitute f ( x ) = 0 in f ( x ) = − 2 3 ( x + 1 2 ) 2 + 10 to compute the x intercept. 0 = − 2 3 ( x + 1 2 ) 2 + 10 − 10 = − 2 3 ( x + 1 2 ) 2 − 30 − 2 = ( x + 1 2 ) 2 15 = ( x + 1 2 ) 2 Simplify further as shown below, x + 1 2 = ± 15 x = − 0

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Chapter 6.5, Problem 23E

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To sketch: The parabola f(x)=−23(x+12)2+10 .

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Chapter 6.5, Problem 22E

Chapter 6.5, Problem 24E

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The parabola f ( x ) = − 2 3 ( x + 1 2 ) 2 + 10 .

Solution Summary: The author illustrates the graph of the parabola f(x)=-23.

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To sketch: The parabola f(x)=−23(x+12)2+10 .

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