Question

please help with #4??

Answer 1

@helpme1.2 ?!

Answer 2

@triciaal

Answer 3

vertex form = a (x-k)+h
so i think
when you plug the given numbers
\[\frac {1}{4}(x-(-8))^2-10\]\[\frac{1}{4}(x+8)^2 -10\]

Answer 4

i got that too.. but thats not standard form!

Answer 5

now you distribute and solve?

Answer 6

standard form would look something like
\[y = ax^2+bx+c\]

Answer 7

yess.. can you get it in that form?

Answer 8

so you have
\[y = \frac{1}{4}(x+8)^2-10 \] expand \[(x+8)^2 \]you would have \[y=\frac{1}{4}(x+8)(x+8)-10 \] now foil \[(x+8)(x+8) \] so you would have \[y=\frac{1}{4}(x^2+8x+8x+64-10 ) \]now combine like terms so it would be \[ y = \frac{1}{4}(x^2+16x-54) \]

Answer 9

cant you keep going?

Answer 10

you can distribute \[\frac{1}{4} \] to \[x^2+16x-54 \] to simplify it even more

Answer 11

i just need it in standard form , can you just get it in standard form???

Answer 12

that is in standard form
if you simplify any farther it would look like
\[y = \frac{1}{4}x+4x-13.5 \] ^^^standard form

Answer 13

thank u very much!!!

Answer 14

no problem