what is the vertex form of y=-1(x-8)^2-59

Question

anonymous · Answer

(8,-59)

anonymous · Answer

how did you get -59 lol i got -51

whpalmer4 · Answer

@sourwing that's the vertex, not to be confused with vertex form...

@prostudent vertex form is $y = a(x-h)^2+k$ with the vertex at $(h,k)$

anonymous · Answer

it's already in the vertex form btw

anonymous · Answer

yeah, i read too fast.

whpalmer4 · Answer

standard form is what you get if you expand it out: 

$$y = ax^2+bx+c$$in this case
$$y = -x^2+16x-123$$

anonymous · Answer

wait so its 8 and -51 for the vertex?

whpalmer4 · Answer

No.  You can read the vertex off directly if you have it in vertex form.  How are you coming up with -51?

whpalmer4 · Answer

Here's the pattern:
$$y = a(x-h)^2+k$$
Here's your formula:
$$y=-1(x-8)^2-59$$
it should be apparent that
$$a=-1$$$$-h=-8$$$$k=-59$$

whpalmer4 · Answer

so the vertex is at $(h,k) = (8,-59)$

anonymous · Answer

o i meant y=-x^2-16x-59      i typed it in wrong :(

whpalmer4 · Answer

So the formula is supposed to be $$y = -x^2-16x-59$$and you want vertex form?  I'd complete the square after factoring out a -1.

$$y = -1(x^2+16x + 59)$$$$y = -1(x^2+16x+64 - 5)$$Can you take it from here?