Vertex.

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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
@jim_thompson5910 If you have time :)
are you able to complete the square for the x terms?

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I'm not sure how.
x^2+12x+_____ what goes in the blank to make that a perfect square?
36
yep
so add and subtract 36 on the same side or add 36 to both sides
then group (x^2-12x+36) and factor that to get (x-6)^2
kks, I chose to add 36 to both sides (x-6)^2+4y+44=36
(x-6)^2+4y=-8 (subtracted 44 from both sides)
now isolate y
y=-(x-6)^2/4 - 8/4
yes
you can simplify that further
2 instead of 8/4 :P
hrm the original and that one don't graph the same o.0
http://www.wolframalpha.com/input/?i=x%5E2%2B12x%2B4y%2B44%3D0 Original
http://www.wolframalpha.com/input/?i=Parabola+y%3D%28%28x-6%29%5E2%29%2F4%29+-2 New equation
Would it work to do: 4y=-x^2-12x-44 y= -x^2/4 -3x -11 ?
Or does the equation we previously made look better?
http://www.wolframalpha.com/input/?i=Parabola+y%3D-1%2F4x%5E2+-3x-11 ....haha what.
let me check
kks
oh I made a mistake, it's not -12x it's +12x
so it should factor to (x+6)^2
you should get \[\Large y = -\frac{1}{4}(x+6)^2 - 2\]
Yep that's what I have :)
that is still a different graph than the original
I have it matching up
oh yeah, sorry. It is. Wolfram was doing a close up of the graph and I hadn't noticed.
you forgot the negative in http://www.wolframalpha.com/input/?i=Parabola+y%3D%28%28x-6%29%5E2%29%2F4%29+-2
ah, sorry.
so next is finding foci, vertex, and directrix.
have a look at this http://www.mathwords.com/f/focus_parabola.htm
k, so mines a vertical parabola.
yes
um and I use x^2=4py to find everything?
Convert \[\Large y = -\frac{1}{4}(x+6)^2 - 2\] to 4p(y-k) = (x-h)^2 form
would that be (y+2)/4=(x+6)^2 ??
more like -4(y+2) = (x+6)^2
-4(y+2) = (x+6)^2 is in the form 4p(y-k) = (x-h)^2 p = -1 h = -6 k = -2
ahh ok
|dw:1433470391325:dw|
|dw:1433470408839:dw|
|dw:1433470431794:dw|
f: = (-6,-3)
yes
d: y = - 3
no
k-p ?
oh sorry, -1
yeah directrix is y = -1
hi can i borrow jim_thompson5910
?
umm so the graph looks like |dw:1433470684289:dw|
|dw:1433470710284:dw|
|dw:1433470757831:dw| ish lol
the prof loves complete accuracy. so if I wanted to find the exact points that the parabola passes though do I just substitute numbers for x into the equation and find out y?
yeah you can do that to plot points or use desmos to get an accurate graph desmos.com/calculator
https://www.desmos.com/calculator
well, i've got to show work. but i'll use desmos to make sure :) thanks!
er it worked until I got to -7 and -8 values for x :o

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