Question

How do you find the vertex for `y=x^2-4`?

Answer 1

`(0, -4)`

Explanation:

To find the vertex of a standard quadratic equation `y = ax^2 + bx + c`, first use the formula `(-b)/(2a)` to find the `x`-coordinate of the vertex:
`x = (-b)/(2a) = (-0)/(2(1)) = 0`

Now plug the `x` value of the vertex back into the equation to find the `y` value:
`y = (0)^2 - 4 = -4`

Therefore, the vertex is at `(0, -4)`.

Hope this helps!