How do you use the important points to sketch the graph of #y= x^2 - 8x + 15#?

1 Answer
Mar 25, 2018

Here you go.

Explanation:

Basically, you need the #x#-intercepts (if any), the #y#-intercepts, and the vertex point. Knowing the concavity helps as well.

To get the #x#-intercept, let #y=0#.

# 0=x^2-8x+15#

#(x-5)(x-3)=0#

So #x# can be #5# or #3# #-># the #x#-intercepts

For the #y#-intercept, I just let #x=0#. This gives the constant, so the #y#-intercept is at the point #(0,15)#.

For the vertex point, I can use #-b/(2a)#, so

#-b/(2a) = 8/2 = 4#

I'm not finished, I need to sub #x=4# back in.

#y=-1# when #x# is #4#.

Now I have the vertex #(4,-1)#, the #x#-intercepts #(3, 0)# and #(5, 0)#, the #y#-intercept #(0,15)#, and the concavity (concave up as it is a positive coefficient of #x^2#).

Now I graph.

graph{x^2-8x+15 [-10, 10, -5, 5]}