# What is the vertex of the parabola y=(x+8)^2+1?

Mar 8, 2016

$\textcolor{b l u e}{{x}_{\text{vertex}} = - 8}$
I have taken you up to appoint where you should be able to finish it off.

#### Explanation:

Standard form $y = a {x}^{2} + b x + c$

Write as:$\text{ } y = a \left({x}^{2} + \frac{b}{a} x\right) + c$

Then ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a}$

Expanding the brackets

$y = {x}^{2} + 16 x + 84 + 1$

In your case $a = 1 \text{ so } \frac{b}{a} = \frac{16}{1}$

Apply $\left(- \frac{1}{2}\right) \times 16 = - 8$

$\textcolor{b l u e}{{x}_{\text{vertex}} = - 8}$

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Find y_("vertex")" " by substitution

$\textcolor{b r o w n}{y = {x}^{2} + 16 x + 85} \textcolor{g r e e n}{\to y = {\left(- 8\right)}^{2} + 16 \left(- 8\right) + 85}$

I will let you finish this bit

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