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

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Answer 1 · 2015-11-20T23:42:35

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

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)#.