x=-5(y+4)^2+3
How do I start
Thanks
Hi ToXic01,
Well, first note that the equation is in vertex form:
$\displaystyle x=a(y-k)^2+h$
This allows you to find the vertex rather quickly.
$\displaystyle V(h, k)$
Since a < 0, the parabola opens to the left.
The axis of symmetry is y = k.
Now, if you want specific points to graph, assign some arbitrary values to y and solve for x.