Need an x^2 somewhere
perhaps
y = 9 (x-8)^2 + 8
or something?
what are the coordinates of the vertex of the parabola defined by:
f(x)=9(x-8)+8
I'm confused on how to solve this. Help pls
3 answers
If it is indeed
y = 9 (x-8)^2 + 8
then
y = 9 (x^2 - 16 x + 64) + 8
y = 9 x^2 - 144 x + 576 + 8
y = 9 x^2 - 144 x + 584
then
x^2 - 16 x + 64 8/9 = y/9
x^2 - 16 x = y/9 -64 -8/9
x^2 - 16 x + 64 = y/9 -64 -8/9 + 64
(x-8)^2 = y/9 -8/9 = (1/9)(y-8)
vertex at (8,8)
y = 9 (x-8)^2 + 8
then
y = 9 (x^2 - 16 x + 64) + 8
y = 9 x^2 - 144 x + 576 + 8
y = 9 x^2 - 144 x + 584
then
x^2 - 16 x + 64 8/9 = y/9
x^2 - 16 x = y/9 -64 -8/9
x^2 - 16 x + 64 = y/9 -64 -8/9 + 64
(x-8)^2 = y/9 -8/9 = (1/9)(y-8)
vertex at (8,8)
of course we could have short cut that by saying
9 (x-8)^2 = y-8
then
(x-8)^2 = (1/9)(y-8)
but best do the whole completing the square process.
9 (x-8)^2 = y-8
then
(x-8)^2 = (1/9)(y-8)
but best do the whole completing the square process.
1 answer