Question

Write the general form of the equation which matches the graph below. In complete sentences, explain the process taken to find this equation.

Answer 1

My lesson describes how to take equations and turn them into parabolas, but it nowhere says how to put a graph in an equation. And I barely understand the first part as it is, so I need help lol. (and i don't want answers. just help)

Answer 2

We're given a parabola with a focus of (3,-4)

This focus is 4 units away from the vertex, so p = 4

Answer 3

The general form of a parabola is

4p(y-k) = (x-h)^2

where (h,k) is the vertex

Answer 4

P? The equation that they give me to write it in is y = a(x – h)^2 + k

Answer 5

So

4p(y-k) = (x-h)^2

4*4(y-k) = (x-h)^2

16(y-k) = (x-h)^2

16(y-0) = (x-3)^2

16y = (x-3)^2

y = (1/16)(x-3)^2

Answer 6

which is now in the form

y = a(x – h)^2 + k

where a = 1/16, h = 3 and k = 0

Answer 7

oh sry, p is the distance (along the axis of symmetry) from the focus to the vertex

Answer 8

does that make sense?

Answer 9

Kind of. So h = -3 and k = 0?

Answer 10

Oh, just saw your other post. It'd be 3 because you have to inverse it. Gotcha

Answer 11

Yeah, I think I understand it now. Thank you

Answer 12

alright that's great :), yw

Answer 13

But could you elaborate on how you got the 16 and turned it to 1/16? I've never been given the equation you used for that

Answer 14

I divided both sides by 16

or equivalently

I multiplied both sides by 1/16

Answer 15

I did that to go from

16y = (x-3)^2

to

y = (1/16)(x-3)^2

Answer 16

okay, thank you!

Answer 17

you're welcome