# How do you find the intercept and vertex of y = x^2 -12?

Dec 24, 2015

The graph of y=x^2 -12 would be a parabola please go through the explanation to understand the approach to work similar problems.

#### Explanation:

Let us understand the vertex form of such parabola.

If the parabola is of form $y = a {\left(x - h\right)}^{2} + k$ then (h,k) would the vertex.

Let us compare our equation with this form.

$y = {x}^{2} - 12$

Let us rewrite this as

$y = 1 {\left(x - 0\right)}^{2} - 12$
$y = a {\left(x - h\right)}^{2} + k$
Comparing we can see $h = 0$ and $k = - 12$

Vertex (0,-12)

Intercepts are the places where the curve cuts the axis. The y intercept is found by equating x=0 and solving for y and the x intercept are found by equating y =0 and solving for x.
In our problem the vertex is the y-intercept as well.

To find x-intercepts equate y=0 and solve.
${x}^{2} - 12 = 0$
${x}^{2} = 12$
$x = \pm \sqrt{12}$
$x = \pm 2 \sqrt{3}$

The x-intercepts are $\left(- 2 \sqrt{3} , 0\right) \mathmr{and} \left(2 \sqrt{3} , 0\right)$

Vertex : $\left(0 , - 12\right)$
y-intercept : $\left(0 , - 12\right)$
x-intercepts : $\left(- 2 \sqrt{3} , 0\right) \mathmr{and} \left(2 \sqrt{3} , 0\right)$