Question

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

Answer 1

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(x-h)^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(x-0)^2 - 12`
`y=a(x-h)^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=+- sqrt(12)`
`x=+- 2sqrt(3)`

The x-intercepts are `(-2sqrt(3),0) and (2sqrt(3),0)`

Final answer
Vertex : `(0,-12)`
y-intercept : `(0,-12)`
x-intercepts : `(-2sqrt(3),0) and (2sqrt(3),0)`