How do you identify the important parts of #x^2-2x-8=0# to graph it?

1 Answer
Rafael
Sep 23, 2015

x-intercepts: #(4,0)# and #(-2,0)#
vertex: #(1,-9)#

Explanation:

I think the most important parts are the x-intercepts and the vertex.

x-intercepts (The value of #x# when #y=0#)
Given the function #y=x^2-2x-8#, set #y# to #0#. Compute for the x-intercepts by factoring.

#y=x^2-2x-8#
#0=x^2-2x-8#
#0=(x-4)(x+2)#

Now that you've factored it, you can say that the x-intercepts are:

#x-4=0#
#x=4#
#color(blue)((4,0)#

#x+2=0#
#x=-2#
#color(blue)((-2,0)#

Mark the points #(4,0)# and #(-2,0)# on your graph.

vertex
To solve for the vertex, you will have to convert the function into the vertex form #y=a(x-h)^2+k#, where #(h,k)# is the vertex.

#y=x^2-2x-8#
#y+8=x^2-2x#
#y+8+1=x^2-2x+1#
#y+9=(x-1)^2#
#y=(x-1)^2-9#

You can see here that #h=1# and #k=-9#. Therefore, your vertex is #(1,-9)#.

After you have plotted the vertex and x-intercepts, just add a few more points by inserting any value to #x#. This will help to make your graph more accurate.

It should look like this:
graph{x^2-2x-8 [-8.97, 11.03, -9.32, 0.68]}

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