Question

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

Answer 1

`color(blue)(x_("vertex")=-8)`
I have taken you up to appoint where you should be able to finish it off.

Explanation:

Standard form `y=ax^2+bx+c`

Write as:`" "y=a(x^2 +b/ax) +c`

Then `x_("vertex")=(-1/2)xxb/a`

Expanding the brackets

`y=x^2+16x +84+1`

In your case `a=1" so "b/a =16/1`

Apply `(-1/2)xx16= -8`

`color(blue)(x_("vertex")=-8)`

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Find `y_("vertex")" "`by substitution

`color(brown)(y=x^2+16x +85)color(green)( -> y=(-8)^2+16(-8) +85)`

I will let you finish this bit

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