What is the vertex of $y = - x^{2} + 40 x - 16$ $y= -x^2+40x-16$ ?

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What is the vertex of $y = - x^{2} + 40 x - 16$ $y= -x^2+40x-16$ ?

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Answer 1 · 2018-08-01T07:24:11

The vertex is at $(20, 384)$ $(20, 384)$ .

Explanation:

Given: $y = - x^{2} + 40 x - 16$ $y = -x^2 + 40x - 16$

This equation is in standard quadratic form $(y = a x^{2} + b x + c)$ $(y = ax^2 + bx + c)$ , meaning we can find the $x$ $x$ -value of the vertex using the formula $\frac{- b}{2 a}$ $(-b)/(2a)$ .

We know that $a = - 1$ $a = -1$ , $b = 4$ $b = 4$ , and $c = - 16$ $c = -16$ , so let's plug them into the formula:
$x = \frac{- 40}{2 (- 1)} = 20$ $x = (-40)/(2(-1)) = 20$

Therefore, the $x$ $x$ -coordinate is $20$ $20$ .

To find the $y$ $y$ -coordinate of the vertex, plug in the $x$ $x$ -coordinate and find $y$ $y$ :
$y = - x^{2} + 40 x - 16$ $y = -x^2 + 40x - 16$

$y = - {(20)}^{2} + 40 (20) - 16$ $y = -(20)^2 + 40(20) - 16$

$y = - 400 + 800 - 16$ $y = -400 + 800 - 16$

$y = 384$ $y = 384$

Therefore, the vertex is at $(20, 384)$ $(20, 384)$ .

Hope this helps!

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What is the vertex of $y = - x^{2} + 40 x - 16$ $y= -x^2+40x-16$ ?

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