SOLUTION: x=5y^2+40y+77 Find the vertex. So far I used (b/2)^2=(40/2)^2=20^2=400 Then x=(5y^2+40y+400)-400+77 Then x=5(y^2+8y+80)-323 I don't know what to do afterwards.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: x=5y^2+40y+77 Find the vertex. So far I used (b/2)^2=(40/2)^2=20^2=400 Then x=(5y^2+40y+400)-400+77 Then x=5(y^2+8y+80)-323 I don't know what to do afterwards.      Log On


   

Question 388163: x=5y^2+40y+77
Find the vertex. So far I used (b/2)^2=(40/2)^2=20^2=400
Then x=(5y^2+40y+400)-400+77
Then x=5(y^2+8y+80)-323 I don't know what to do afterwards.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Factor out the 5 first.
x=5y%5E2%2B40y%2B77
x=5%28y%5E2%2B8y%29%2B77
x=5%28y%5E2%2B8y%2B16%29%2B77-5%2816%29
x=5%28y%2B4%29%5E2%2B77-90
highlight%28x=5%28y%2B4%29%5E2-3%29