How do you find the vertex and intercepts for y = x^2 - 8?

1 Answer
Nov 20, 2015

vertex = (0, -8)
y-intercept = -8
x-intercepts = 2sqrt(2) and -2sqrt(2)

Explanation:

The general equation for a quadratic function in vertex form is:

y = a(x-h)^2+k

With this equation, y=x^2-8 can be rewritten as:

y=1(x-0)^2-8

The vertex of a quadratic equation in vertex form is (h, k), or in this case (0, -8).

To find the y-intercept, substitute x as 0, since the x-coordinate of the y-intercept is 0:

y=x^2-8
y=(0)^2-8
y=(0)-8
y=-8

To find the x-intercept, substitute y as 0, since the y-coordinate of the x-intercept is 0:

y=x^2-8
0=x^2-8
8=x^2
x=+-sqrt(8)
x=2sqrt(2) or -2sqrt(2)
x=2.83 or -2.83

Here is a graph of the equation:
graph{y= x^2-8 [-12.78, 12.53, -11.14, 2.05]}

As you can see, the graph has a y-intercept of -8 and x-intercepts of 2sqrt(2) and -2sqrt(2).