We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
107
3
avatar

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
avatar
0

I have been stuck on this for a long time. 

 Aug 11, 2019
 #2
avatar+105538 
+2

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
avatar
+1

Thx for the help!! laugh

 Aug 12, 2019

10 Online Users

avatar