# Math Help - Graphing Parabola

1. ## [Solved] Graphing Parabola

x=-5(y+4)^2+3

How do I start

Thanks

2. Originally Posted by ToXic01
x=-5(y+4)^2+3

How do I start

Thanks
Hi ToXic01,

Well, first note that the equation is in vertex form:

$x=a(y-k)^2+h$

This allows you to find the vertex rather quickly.

$V(h, k)$

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.

3. how about focus, directrix,

4. Originally Posted by ToXic01
how about focus, directrix,
You'll find the focus at $\left(h+\frac{1}{4a}, k\right)$

and the directrix at $x=h-\frac{1}{4a}$

5. You're welcome.

6. Thanks