What is the vertex of $y = 2 {(x + 3)}^{2} - 8 x$ $y=2(x +3)^2 -8x$ ?

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What is the vertex of $y = 2 {(x + 3)}^{2} - 8 x$ $y=2(x +3)^2 -8x$ ?

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Answer 1 · 2015-12-21T16:10:26

The vertex is $(- 1, 16)$ $(-1,16)$ .

Explanation:

In order to know, we will develop first, it will make the next calculus easier.

$y = 2 x^{2} + 12 x + 18 - 8 x = 2 x^{2} + 4 x + 18$ $y = 2x^2 + 12x + 18 - 8x = 2x^2 + 4x + 18$ . The coefficient of $x^{2}$ $x^2$ is positive so we know the vertex is a minimum. This vertex will be the zero of the derivative of this trinomial. So we need its derivative.

$f (x) = 2 x^{2} + 4 x + 18$ $f(x) = 2x^2 + 4x + 18$ so $f'(x) = 4x + 4 = 4(x+1)$ . This derivative is zero for $x = -1$ so the vertex is at the point $(-1,f(-1)) = (-1,16)$

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What is the vertex of $y = 2 {(x + 3)}^{2} - 8 x$ $y=2(x +3)^2 -8x$ ?

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Explanation:

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Science

Math

Humanities

... and beyond

What is the vertex of y=2(x +3)^2 -8x y=2(x+3)2−8xy=2(x +3)^2 -8x ?

1 Answer

Explanation:

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What is the vertex of $y = 2 {(x + 3)}^{2} - 8 x$ $y=2(x +3)^2 -8x$ ?