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.

Guest Aug 11, 2019

#2**+1 **

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.

Melody Aug 11, 2019