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What is the maximum or minimum value of the function? What is the range?


y = –2x2 + 32x –12 (1 point)

maximum: 116

range: y 116

maximum: –116

range: y –116

maximum: 116

range: y 116

maximum: –116

range: y –116

Verified Answer :

You can use the formula
f(x) = ax^2+bx+c
f(x) = a(x-xv)^2+yv
where xv = -b/2a = -32/-4=8
yv = f(xv) = 116

f(x) = -2(x-8)^2+116

The maximum point is (8, 116).  The answer to your question is C. I hope that this is the answer that you were looking for and it has helped you.

The correct answer is:

maximum: 116
; range: y ≤ 116

Explanation:

This is a quadratic equation in standard form, which is y=ax²+bx+c. The maximum or minimum of a quadratic function is the vertex. To find the x-coordinate of the vertex, we find the axis of symmetry. This is given by the formula x=-b/2a:

x = -32/2(-2) = -32/-4 = 8

To find the y-coordinate, plug this into the equation:

y = -2(8²)+32(8)-12

y=-2(64) + 256 - 12 = -128+256-12 = 128-12 = 116

The coordinates of the vertex are (8, 116).

To determine if this is a maximum or minimum, look at the value of a. It is -2. Since it is negative, this means the parabola opens downward, and the vertex is a maximum.

Since this is a maximum at y=116, this means the range, our y-values, will be less than or equal to this value of 116.

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