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

Dec 12, 2017

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:
$- \frac{b}{2 a}$ for a quadratic equation $a {x}^{2} + b x + c$

Therefore, for vertex $\left(x , y\right)$,
$x = \setminus \frac{- 0}{2 \cdot 1} = 0$

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

Therefore, the vertex is $\left(0 , - 4\right)$.

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 $y = a {\left(x - p\right)}^{2} + q$ where the vertex is $\left(p , q\right)$.

However, the equation $y = {x}^{2} - 4$ is already in vertex form (This is equal to $y = 1 {\left(x - 0\right)}^{2} + \left(- 4\right)$.

Therefore, the coordinates of the vertex is $\left(0 , - 4\right)$.

Hope that makes sense!