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Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,1). Express your answer in the form "ax^2+bx+c".

 

This should be very easy but I can't figure it out. Any help will be great.

 Aug 11, 2019
 #1
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I have been stuck on this for a long time. 

 Aug 11, 2019
 #2
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Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,1). Express your answer in the form "ax^2+bx+c".

 

\((x-2)^2=4a(y-4)\\ sub\; in\; (1,1)\\ 1=4a*-3\\ a=\frac{-1}{12}\\ so\\ (x-2)^2=\frac{-(y-4)}{3}\\ \)

 

Now rearrange it to get the form you want.

 

I checked it with Desmos, which is something you should always do too.

 

https://www.desmos.com/calculator/rg9eqzxyxp

 Aug 11, 2019
 #3
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+1

Thx for the help!! laugh

 Aug 12, 2019

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