How do you graph #f(x)=(x-6)^2+3# and identify the x intercepts, vertex?

Question

5387 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-27T21:27:48

See below

The quadratic is already in vertex form:

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

So, you can identify the vertex by inspection:

#(h,k)=(6,3)#

To find the #x#-intercepts, set #y=0#:

#(x-6)^2+3=0#

#(x-6)^2=-3#

#x-6=sqrt(-3)#

#x=6+-sqrt(-3)#

#x=6+3i#
#x=6-3i#

As you can see, the #x#-intercepts are imaginary.

graph{(x-6)^2+3 [-5, 10, -2, 8]}