Question

what is the equation of the following graph?

Answer 1

It opens to the right, so we know that it will be in the form:
x = a(y-k)²+h
Where (h,k) is the focus and a is positive.

Answer 2

so how do i know what numbers to replace the variables with

Answer 3

So the focus is at (h,k)
The focus is the point that is the furthest left in this graph.

Answer 4

(1, 2) ?

Answer 5

Yup.
Now, a is how squished/stretched the graph is.

From the focus, how many points up do you have to go before the graph has moved one point left?

Answer 6

*one point right

Answer 7

3?

Answer 8

Yes. So that means it's squished to ⅓ of normal.

Your final function is
x = ⅓(y-2)² + 1

Answer 9

that just doesn't seem to match up with a different graph. like when i graphed y = 1/32x^2 i got that the focus was (0, 8). is that right?

Answer 10

@DDCamp

Answer 11

\[\frac{ 1 }{ 32 }x^2?\]

Answer 12

yup

Answer 13

[Facepalm] I meant to say vertex, not focus.

Answer 14

hahahah that's what i figured.

Answer 15

@DDCamp can you help me with this one real quick? i have to write the equation for this graph too. sorry, i just wanna make sure i'm doing it right.

Answer 16

It opens up, so we know it uses the form:
y = a(x-h)² + k

Can you find the vertex, (h,k)?

Answer 17

y = a(x -1)^2 + 4

Answer 18

Yes. Can you find how squished/stretched it is?

Answer 19

2 or -2?

Answer 20

Since it opens up, we know that a is positive.
But since there is a run of 2 with a rise of 1, the factor is ½

Answer 21

so the final equation is y = 1/2(x-1)^2 + 4 ?

Answer 22

Yes

Answer 23

thank you