How do you convert $y = x^{2} + 18 x + 95$ $y=x^2+18x+95$ in vertex form?

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How do you convert $y = x^{2} + 18 x + 95$ $y=x^2+18x+95$ in vertex form?

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Answer 1 · 2018-03-15T15:22:27

$y = {(x + 9)}^{2} + 14$ $y=(x+9)^2+14$

Explanation:

First find the vertex using the formula
$x = \frac{- b}{2a}$ $x=(-b)/"2a"$

$a = 1$ $a=1$
$b = 18$ $b=18$
$c = 95$ $c=95$

$x = \frac{- (18)}{2(1)}$ $x=(-(18))/"2(1)"$ This simplifies to $x = - \frac{18}{2}$ $x=-18/"2"$ which is $- 9$ $-9$ .
so $x = - 9$ $x=-9$

So on now that we have $x$ $x$ we can find $y$ $y$ .

$y = x^{2} + 18 x + 95$ $y=x^2+18x+95$
$y = {(- 9)}^{2} + 18 (- 9) + 95$ $y=(-9)^2+18(-9)+95$
$y = 14$ $y=14$

Vertex = $(- 9, 14)$ $(-9,14)$ where $h = - 9$ $h=-9$ and $k = 14$ $k=14$

We now finally enter this into vertex form which is,
$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

$x$ $x$ and $y$ $y$ in the "vertex form" are not associated with the values we found earlier.

$y = 1 {(x - (- 9))}^{2} + 14$ $y=1(x-(-9))^2+14$
$y = {(x + 9)}^{2} + 14$ $y=(x+9)^2+14$

Answer 2 · 2018-03-15T15:25:36

${(x + 9)}^{2} + 14$ $(x+9)^2+14$

Explanation:

Firstly, you have to find the vertex point. Use the formula $x_{v} = - \frac{b}{2 a}$ $x_v=-b/(2a)$ . You get $(h) x_{v} = - 9$ $(h) x_v=-9$ and $(k) y_{v} = 14$ $(k)y_v=14$

Since there's an imaginary 1 in front of x, the a value is 1.
Now just plug everything into the equation $y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

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How do you convert $y = x^{2} + 18 x + 95$ $y=x^2+18x+95$ in vertex form?

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How do you convert y=x^2+18x+95y=x2+18x+95y=x^2+18x+95 in vertex form?

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How do you convert $y = x^{2} + 18 x + 95$ $y=x^2+18x+95$ in vertex form?