How do you find the maximum or minimum value of $y = 2 x^{2} + 32 x - 4$ $y=2x^2 +32x - 4$ ?

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How do you find the maximum or minimum value of $y = 2 x^{2} + 32 x - 4$ $y=2x^2 +32x - 4$ ?

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Answer 1 · 2017-08-13T08:50:54

Find the derivative, $\frac{d y}{d x}$ $dy/dx$ , and solve for $\frac{d y}{d x} = 0$ $dy/dx =0$ .
The $x$ $x$ -coordinate of the turning point is at $x = - 8$ $x=-8$ .

Explanation:

The derivative of any term $x^{n}$ $x^n$ is given by $n x^{n - 1}$ $nx^(n-1)$ , so for this polynomial, $2 x^{2} + 32 x - 4$ $2x^2+32x-4$ , the derivative is:

$2 \cdot 2 x + 1 \cdot 32 + 0 \cdot - 4$ $2*2x + 1*32 + 0*-4$ .
$:. dy/dx = 4x +32$ .

To calculate the turning point(s), let this equal $0$ .

$4x + 32 = 0$
$:. x= -8$ .

By substituting this into the original equation, the turning point is at $(-8, -137)$

Since $dy/dy > 0$ for $x = -7$ and $dy/dx < 0$ for $x = -9$ , this is a minimum turning point.

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How do you find the maximum or minimum value of $y = 2 x^{2} + 32 x - 4$ $y=2x^2 +32x - 4$ ?

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How do you find the maximum or minimum value of y=2x^2 +32x - 4y=2x2+32x−4y=2x^2 +32x - 4?

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How do you find the maximum or minimum value of $y = 2 x^{2} + 32 x - 4$ $y=2x^2 +32x - 4$ ?