You can put this solution on YOUR website! Find the vertex and focus of y=x^2+6x+7 and sketch it's graph.
Y=X^2+2*X*3+3^2-3^2+7=(X+3)^2-2
Y+2=1(X+3)^2
COMPARING WITH STD.EQN.OF PARABOLA
(X-H)^2=4A(Y-K)..WHOSE VERTEX IS (H,K)AND FOCUS IS (H,K+A)
(X+3)^2=(Y+2)
WE FIND H=-3...K=-2......A=1/4.......HENCE
VERTEX IS (-3,-2) AND
FOCUS IS (-3,-2+1/4)=(-3,-7/4)
GRAPH IS
