What is the maximum or minimum value of the function, what is the range y = -2x2 + 32x - 12?
Solution:
Given, the function is y = -2x2 + 32x - 12
We have to find the maximum or minimum value of the function.
The maximum or minimum of a quadratic function occurs at x = -b/2a.
If a is negative, the maximum value of the function is f(-b/2a).
If a is positive, the minimum value of the function is f(-b/2a).
\(f_{max}x=ax^{2}+bx+c\) occurs at x = -b/2a. --- (1)
From the given equation,
a = -2, b = 32, c = -12
Substitute the value of a and b in (1)
x = -32/2(-2)
x = -32/-4
x = 8
Put x = 8 in the given equation,
y = -2(8)2 + 32(8) - 12
y = -2(64) + 256 - 12
y = -128 + 256 - 12
y = 256 - 140
y = 116
From the graph, it is clear that the given function represents a parabola that opens downwards with vertex (8, 116).
The range of the function extends to negative infinity.
Therefore, the range of the function lies in the interval \((-\infty , 116)\).
What is the maximum or minimum value of the function, what is the range y = -2x2 + 32x - 12
Summary:
The maximum or minimum value of the function y = -2x2 + 32x - 12 is (8, 116) and the range of the function y = -2x2 + 32x - 12 lies in the interval \((-\infty , 116)\).
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