consider the following quadratic function...k(x)=(x-6)..find the vertex of this function

Question

anonymous · Answer

that's not a quadratic function

anonymous · Answer

is (x-6) raised to the 2nd power?

anonymous · Answer

no...it's not on the problem...that's how it's worded

anonymous · Answer

k(x)=x-6 is a line

anonymous · Answer

I have the answer..I just don't understand how to do the problem...the answer  is (6,0)

anonymous · Answer

Well, I'm going to assume it is k(x)=(x-6)^2
Expand: x^2-12x+36
The vertex of a parabola written in the for ax^2+bx+c is:
$$(-\frac{b}{2a},f(-\frac{b}{2a})$$
$$(\frac{12}{2},f(\frac{12}{2}))=(6,f(6))=(6,0)$$

anonymous · Answer

looks good....worked it out..now quick question...why is f(6)=0?

anonymous · Answer

f(6)=6^2-12(6)+36=36-72+36=0

anonymous · Answer

or in the original form,
f(x)=(x-6)^2
f(6)=(6-6)^2=0^2=0

anonymous · Answer

makes sense now thank you