Question

Vertex.

Answer 1

Write the equation in standard form then find and list the vertex, focus, and directrix of the parabola. sketch its graph showing the focus and the directrix.(note: Choose the locations of the coordinate axes and draw them on the grid and choose scales so that your graph is, if possible, about three by three inches in size.)

x^2+12x+4y+44=0

Answer 2

@jim_thompson5910 If you have time :)

Answer 3

are you able to complete the square for the x terms?

Answer 4

I'm not sure how.

Answer 5

x^2+12x+_____
what goes in the blank to make that a perfect square?

Answer 6

36

Answer 7

yep

Answer 8

so add and subtract 36 on the same side

or

add 36 to both sides

Answer 9

then group (x^2-12x+36) and factor that to get (x-6)^2

Answer 10

kks, I chose to add 36 to both sides
(x-6)^2+4y+44=36

Answer 11

(x-6)^2+4y=-8
(subtracted 44 from both sides)

Answer 12

now isolate y

Answer 13

y=-(x-6)^2/4 - 8/4

Answer 14

yes

Answer 15

you can simplify that further

Answer 16

2 instead of 8/4 :P

Answer 17

hrm the original and that one don't graph the same o.0

Answer 18

http://www.wolframalpha.com/input/?i=x%5E2%2B12x%2B4y%2B44%3D0 Original

Answer 19

http://www.wolframalpha.com/input/?i=Parabola+y%3D%28%28x-6%29%5E2%29%2F4%29+-2 New equation

Answer 20

Would it work to do:
4y=-x^2-12x-44
y= -x^2/4 -3x -11 ?

Answer 21

Or does the equation we previously made look better?

Answer 22

http://www.wolframalpha.com/input/?i=Parabola+y%3D-1%2F4x%5E2+-3x-11 ....haha what.

Answer 23

let me check

Answer 24

kks

Answer 25

oh I made a mistake, it's not -12x it's +12x

Answer 26

so it should factor to (x+6)^2

Answer 27

you should get
\[\Large y = -\frac{1}{4}(x+6)^2 - 2\]

Answer 28

Yep that's what I have :)

Answer 29

that is still a different graph than the original

Answer 30

I have it matching up

Answer 31

oh yeah, sorry. It is. Wolfram was doing a close up of the graph and I hadn't noticed.

Answer 32

you forgot the negative in

http://www.wolframalpha.com/input/?i=Parabola+y%3D%28%28x-6%29%5E2%29%2F4%29+-2

Answer 33

ah, sorry.

Answer 34

so next is finding foci, vertex, and directrix.

Answer 35

have a look at this

http://www.mathwords.com/f/focus_parabola.htm

Answer 36

k, so mines a vertical parabola.

Answer 37

yes

Answer 38

um and I use x^2=4py to find everything?

Answer 39

Convert \[\Large y = -\frac{1}{4}(x+6)^2 - 2\] to 4p(y-k) = (x-h)^2 form

Answer 40

would that be
(y+2)/4=(x+6)^2 ??

Answer 41

more like -4(y+2) = (x+6)^2

Answer 42

-4(y+2) = (x+6)^2 is in the form 4p(y-k) = (x-h)^2

p = -1
h = -6
k = -2

Answer 43

ahh ok