# What is the the vertex of y = -x^2 - 3?

Jan 5, 2016

$V e r t e x : \left(0 , - 3\right)$

#### Explanation:

$y = - {x}^{2} - 3$
Let us first convert this in vertex from

$\textcolor{b r o w n}{\text{vertex form:y=a(x-h)^2+k}}$
$\textcolor{b r o w n}{\text{vetex:(h,k)}}$
Let us write the given equation in vertex form.
$y = {\left(x - 0\right)}^{2} + \left(- 3\right)$
$V e r t e x : \left(0 , - 3\right)$

Jan 5, 2016

$\text{vertex} \to \left(x , y\right) \to \left(0 , - 3\right)$

Explanation shows what is happening.

#### Explanation:

Suppose we hade the general equation of ${y}_{1} = - {x}^{2}$

Then the graph would look like:

Subtract 3 from both sides of the equation. Not only is the equation now ${y}_{1} - 3 = - {x}^{3} - 3$ but you have lowered the whole thing by 3.
Let ${y}_{1} - 3$ be written as ${y}_{2}$ now giving: ${y}_{2} = {x}^{2} - 3$
This graph looks like:

From this you can see that the vertex in the $\textcolor{b l u e}{\text{first case}}$ is at ${x}_{\text{vertex")=0" and "y_("vertex}} = 0$ written as $\text{vertex} \to \left(x , y\right) \to \left(0 , 0\right)$

In the $\textcolor{b l u e}{\text{second case}}$ it has lowered by 3 on the x-axis giving ${x}_{\text{vertex")=0" and "y_("vertex}} = - 3$ written as

$\text{vertex} \to \left(x , y\right) \to \left(0 , - 3\right)$