Question

Find the vertex, focus, and directrix of the parabola and sketch its graph. Use a graphing utility to verify your graph.
y2 − 8y − 4x = −12

Answer 1

You don't have to graph it for me
just explain to me how to get the vertex because i keep getting the wrong answer :-(

Answer 2

do you know how to complete the square ?

Answer 3

like y^2-8y+?
what would the question mark need to be to make y^2-8y+? a square

Answer 4

16

Answer 5

right
\[y^2+ky+(\frac{k}{2})^2=(y+\frac{k}{2})^2 \\ y^2-8y+(\frac{8}{2})^2=....\]
so y^2-8y+16=?

Answer 6

4x-12

Answer 7

...we aren't looking at your equation right now

Answer 8

I'm asking you to write y^2-8y+16 as a square right now

Answer 9

(y-4)^2

Answer 10

ok cool stuff

let's look at your equation now
so we need to add 16 to both sides of our equation

Answer 11

so that we can write that (y-4)^2 somewhere on the left hand side

Answer 12

\[y^2-8y-4x=-12 \\ y^2-8y+16-4x=-12+16 \\ \\ \text{ and it looks like you were wanting to go ahead and add } 4x \\ \text{ \to both sides } \\ \text{ \let's do that } \\ y^2-8y+16=4x-12+16 \\ (y-4)^2=4x-12+16\]

Answer 13

so anyways you can simplify -12+16

Answer 14

and factor whatever coefficient of x is out of the terms on the right hand side

Answer 15

(y-4)^2=4(x+1) is the equation then?

Answer 16

right

Answer 17

thanks

Answer 18

so you are able to spot the vertex?

Answer 19

(-1, 4)

Answer 20

ok cool
that's it

Answer 21

i didn't know we had to add 16 to both sides so that's where i was going wrong oops

Answer 22

yeah in order for equation to remain the same equation you must do to one side as you would do to the other

Answer 23

like for example you know x=x is true
but if you add 16 to one side then the equation is false x+16=x is not true

Answer 24

that makes sense now

Answer 25

but wee know that if x=x is true then x+16=x+16 is true :)

Answer 26

x+16=x+16 is true anyways
this was just an example

Answer 27

here is another
if x=y is true, then x+16=y+16 is true

Answer 28

anyways good luck