How do you find the vertex and intercepts for y=x2−8?

1 Answer
Johnson Z.
Nov 20, 2015

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

Explanation:

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

y=a(x−h)2+k

With this equation, y=x2−8 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=x2−8
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=x2−8
0=x2−8
8=x2
x=±√8
x=2√2 or −2√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 2√2 and −2√2.

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
3315 views around the world
You can reuse this answer
Creative Commons License