Question

What is the vertex of `y= (x-2)^2-3x^2-4x-4`?

Answer 1

`(-2,8)`

Explanation:

The formula for the x-value of the vertex of a quadratic is:

`(-b)/(2a)="x-value of the vertex"`

To get our `a` and `b`, it's easiest to have your quadratic in standard form, and to get that, work your quadratic all the way out and simplify, getting you:

`y=x^2-4x+4-3x^2-4x-4`

`y=-2x^2-8x`

In this case, you have no `c` term, but it doesn't really affect anything. Plug in your `a` and `b` into the vertex formula:

`(-(-8))/(2(-2))="x-value of the vertex"`

`"x-value of the vertex"=-2`

Now plug your newly found `"x-value"` back into your quadratic to solve for its `"y-value"`, which gives you:

`y=-2(-2)^2-8(-2)`

`y=8`

Concluding that the coordinates of the vertex of this quadratic are:

`(-2,8)`