Question

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

Answer 1

I know this question is one year old....

But for the others who may want to know how to do this, here is the solution:

To find vertexes, there are two methods.

Method one: Using the formula

You can easily find the `x`-coordinate of the vertex by using the formula:
`-b/(2a)` for a quadratic equation `ax^2 + bx + c`

Therefore, for vertex `(x, y)`,
`x = \frac{-0}{2 * 1} = 0`

Then, you calculate the `y`-coordinate with the given equation:
`y = x^2 - 4 = 0^2 - 4 = -4`

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

Method two: Completing the square

This method is more formal and some tests and exams require you to use this method.

This method will find the vertex form `y = a(x - p)^2 + q` where the vertex is `(p, q)`.

However, the equation `y = x^2 - 4` is already in vertex form (This is equal to `y = 1(x - 0)^2 + (-4)`.

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

Hope that makes sense!