# How do you find the vertex and the intercepts for G(x) = x ^ 2 - 6?

Apr 11, 2016

$x$ intercepts at: $x = \pm \sqrt{6}$
$y$ intercepts at the point $\left(0 , - 6\right)$.
Vertex at the point $\left(0 , - 6\right)$

#### Explanation:

Let's first have a look at the $x$ intercepts of the function:

Recall that the difference between two squares can be expressed as: ${n}^{2} - {m}^{2} = \left(n - m\right) \left(n + m\right)$

If we consider 6 as:

${\sqrt{6}}^{2}$
We can now convert the function to a difference of squares:
$G \left(x\right) = {x}^{2} - 6$
$G \left(x\right) = \left(x - \sqrt{6}\right) \left(x + \sqrt{6}\right)$

Using the Null Factor Law we can now solve for $x$, hence determine the $x$ intercepts of the function:

$0 = \left(x - \sqrt{6}\right) \left(x + \sqrt{6}\right)$

Therefore $x$ intercepts are present at:
$x = \pm \sqrt{6}$

Let's now have a look at the $y$ intercept of the function:

Consider that for any function, $y$ intercepts where $x = 0$. Therefore if we substitute $x = 0$ into the function, we can calculate the $y$ intercept of the function.

$G \left(x\right) = {x}^{2} - 6$
$G \left(x\right) = {0}^{2} - 6$
$G \left(x\right) = - 6$

Therefore, the $y$ intercept of the function is present at the point $\left(0 , - 6\right)$.

Finally, let's have a look at the vertex of the function:

Consider the general form of a parabolic function:
$y = a {x}^{2} + b x + c$

If we compare the equation that you have presented:

$G \left(x\right) = {x}^{2} - 6$

We can determine that:

The ${x}^{2}$ coefficient is 1; this implies that $a = 1$

The $x$ coefficient is $0$; this implies that b = 0
The constant term is $- 6$; this implies that
c = -6
Therefore, we can use the formula:
Tp_x=−b/2a

to determine the $x$ value of the turning point.

Substituting the appropriate values into the formula we get:

Tp_x=−0/(2*1)
Tp_x=−0/(2)
$T {p}_{x} = 0$

Substituting this value into the given function, we get:

$G \left(x\right) = {0}^{2} - 6$
$G \left(x\right) = {0}^{2} - 6$
$G \left(x\right) = - 6$

Therefore, the vertex of the function is present at the point $\left(0 , - 6\right)$.