Question

Y=x^2-18x-59
Convert into vertex form.
Thanks

Answer 1

http://www.mathwarehouse.com/geometry/parabola/images/standard-vertex-forms-thumb.png

Answer 2

I need to do it step by step

Answer 3

ok

Answer 4

Im struggling with the formula

Answer 5

is just a matter of "completing the square" for a "perfect square trinomial"
I'd think you've covered this but anyhow

a perfect square trinomial looks like -> \(\large x^2\pm2xy+y^2 \implies (x\pm y)^2 \)

Answer 6

Y=(x^2-18x )-59
Y=(x^2-18x+81)-59
y=X^2 -18X+81-81)-59

Answer 7

that's correct :)

Answer 8

so you have \( (x-9)^2 -81-59\)

Answer 9

I get lost at this point

Answer 10

thats the answer I get but the back of the book says the answer is x^2+6x=9

Answer 11

Y=(x^2-18x )-59
Y=(x^2-18x+81)-59
y=X^2 -18X+81-81)-59

so you have
$$
y=(x^2-18x+81)-81-59\\
y=(\color{red}{x}^2-2(\color{blue}{9})(\color{red}{x})+\color{blue}{9}^2)-81-59\\
y=(\color{red}{x}-\color{blue}{9})^2-81-59
$$

Answer 12

sorry x^2-4x+4

Answer 13

hmm, that doesn't like it'd be it :/

Answer 14

what the above gives is (x-9)^2 -140

Answer 15

you can always recheck that in your calculator, plot the equation, check where the vertex (h,k) values are at

Answer 16

Thanks for the help. This example help out alot

Answer 17

yw