the equation of a parabola is 12y=(x-1)^2-48. Identify the vertex focus and the directrix of the parabola
You still there?
yes
@legomyego180
this is the last question on my quiz i really need help
Are you familiar with vertex form?
uhh not really this is really confusing to me
It's ok. To find your vertex when the equation is written like this look at whatever x is being subtracted by. Can you tell what it is in this case?
1?
Keep in mind this only works when the equation is in this form: \[y=(x-a)^2-c\] where a and c are any number. Yes, 1. This means that the x-value of your vertex is one, so we have figured out half of it. (1,y) Still need to find the y value. Any ideas on how to do that?
divide 12 y from each side?
Sure, you can do that. But if you have an x value, and you want to find its corresponding y vaule, you can just plug it back into the function. So plug x=1 into your function and see what you get for y. That will give you both points of your vertex. Does that make sense?
yes ok let me try that
i got 4
-4
oh ok i probably missed a step or something
so thats the vertex?
(1,-4) = vertex. Keep in mind if this was something like (x+1) instead of (x-1) your x value for the vertex would be -1. Its whatever makes the thing inside the parentheeses zero.
ok got it
what about the rest
One sec. I havent done this in a while, so I'm learning this as I teach it
ok
I reccomend watching that all the way through.
ok thanks
So first thing lets rewrite \[12y=(x-1)^2-48\] as \[y+48=\frac{ 1 }{ 12 }(x-1)^2\] because w want to get the x^2 value by itself.
Actually, that should just be \[12y+48=(x-1)^2\]
Then: 4P=12 P=3 So add 3 to your vertex's y value to get your Focus, and subtract P to get your directrix. Sorry, im kind of lousy at this stuff
But it is the correct answer.
thanks its ok!
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