Question

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

Answer 1

vertex`-> (x,y)-> (-8,-2)`

Explanation:

When a quadratic is in this from `x_("vertex") = (-1) xx b`

where `b-> (x+b)^2`

In truth, if the original equation was of form:
`y=ax^2+b+c`..............................(1)

and `k` is a corrective value and you write equation (1) as:

`y=a(x+b/a)^2+k+c`

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

However, in your case, `a=1`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`x_("vertex")= (-1)xx8 =-8`

Having found this just substitute into the original equation to find the value of `y_("vertex")`

So we have: `y=( (-8)+8)^2-2" "=" "-2`

so the vertex`-> (x,y)-> (-8,-2)`

Answer 2

( -8 , -2 )

Explanation:

The equation of a parabola in vertex form is :

`y = (x - h)^2 + k`
where (h , k ) are the coords of the vertex.

here `y = (x +8 )^2 -2`

and by comparison h = -8 and k = -2 → vertex = (-8 , -2 )
graph{(x+8)^2-2 [-10, 10, -5, 5]}