Sketch the graph y^2 + 12x = 2y - 13 Find its vertex.
We have given
shift the variables and we have
make lhs perfect square so ,we can write
Which is equation of parabola and defined if
parabola is open left i.e second and third quadrant .
`vertex` is (-1,1)
What have we to do with?
`x=0 ` not for real value of `y`
explicit function `y=y(x)` defined for `x>=1`
We note function hasn't critical points, indeed:
that never will be like 0.
Never if we concern about original function:
we wl lfaound critcal points:
Indeed: `(del f)/(del y) =2(y-1)` zero for `y=1`
but `(del f)/(del x)=-12` that won'tever be zero.