Question

Find the vertex of the parabola
X^2-4x-4y+16=0

Answer 1

ok... so rewrite the equation as

4y = (x^2 -4x) + 16

complete the square in x as the 1st step.

Answer 2

what do you get...?

Answer 3

Y=(x^2+x)+4

Answer 4

no you need to complete the square so

\[4y = (x^2 -4x + 4) + 12\]

now factor

\[4y = (x -2)^2 + 12\]

divide everything by y...

\[y^2= \frac{1}{4}(x -2)^2 + 3\]

this is one vertex form of the equation

the general form is

\[y = a(x - h)^2 + k\]

where the vertex is at (h, k)

match you question to the general form to find the vertex

Answer 5

So the vertex is (2,3)

Answer 6

that's correct.

Answer 7

Thank you