# What is the vertex of  y=x^2-4 ?

Mar 4, 2018

$V e r t e x \left(0 , - 4\right)$

#### Explanation:

.

$y = {x}^{2} - 4$

If the equation of a parabola is in the form:

$y = a {x}^{2} + b x + c$

we can find the $x$-coordinate of its vertex using the following formula:

${x}_{v e r t e x} = - \frac{b}{2 a}$

Comparing the problem equation with the form above, we see:

$a = 1 , b = 0 , c = - 4$

${x}_{v e r t e x} = - \frac{0}{2 \left(1\right)} = 0$

Now, we can plug this into the equation to find the $y$-coordinate:

${y}_{v e r t e x} = {\left(0\right)}^{2} - 4 = 0 - 4 = - 4$

Therefore,

$V e r t e x \left(0 , - 4\right)$

You can see the graph of this parabola below:

graph{x^2-4 [-10, 10, -5, 5]}