Question

Find the Vertex, Focus, Directrix and Graph

Answer 1

\[x ^{2}+8x=4y-8\]

I found the Vertex and everything but its telling me i am wrong. @campbell_st

Answer 2

what did you get for the vertex?

Answer 3

Umm let me show you my steps for better understanding
x2+8x=4y-8

(x+4)^2=4a(y-8)
V = (-4,8)
D = 1

Answer 4

give me a sec to find the mistake
the \(-4\) is right

Answer 5

Yea they told me the 8 is wrong D:

Answer 6

oh i see it
when you complete the square on the left to turn \(x^2+8x\) in to \((x+4)^2\) you have to add \(4^2=16\) to the right

Answer 7

\[x^2+8x=4y-8\\
(x+4)^2=4y-8+16\] is a start

Answer 8

you have an error in the factoring of the right hand side...

Answer 9

if you use the model

\[(x - h)^2 = 4a(y - k)\]

the vertex is (h, k) the focal length is a

from there you can find the focus and directrix

Answer 10

So @satellite73
(x+4)^2 = 4a(y+4) ?

Answer 11

now... if you have 4y - 8 and take 4 asa common factor what's left?

Answer 12

forget the \(a\) \[(x+4)^2=4y+8\] after you add the \(16\) the factor out the four \[(x+4)^2=4(y+2)\]

Answer 13

sorry my mistake...

Answer 14

OHH yeaaaa How did i forget about factoring D:

Answer 15

start by completing the square on the left... and adding the same value to the right

Answer 16

Ah so when we divide the left side by 2 we square it again and add it to the right side ?

Answer 17

there is no dividing by two

Answer 18

\[x^2+8x\] on the left
forget about the right

Answer 19

Oh i mean factor

Answer 20

when you complete the square you have to add 16 to the other side \[(x+4)^2=x^2+8x+16\] so if you add 16 to the left you have to add 16 to the right
that turns \[4y-8\] in to \[4y+8\] on the left hand side

Answer 21

Yes i get that

Answer 22

i meant "on the right"
so you have \[(x+4)^2=4y+8\] factor out the 4 on the right

Answer 23

So we get
4(y+2)

Answer 24

yes

Answer 25

Ah i get it so the A = 1

Answer 26

now you can read off your answer from the standard form \[(x+4)^2=4(y+2)\]