Graphing a Parabola with Vertex (h, k) and Axis of Symmetry Parallel to the x-axis

The standard form that applies to the given equation is ${\left(y-k\right)}^2=4p\left(x-h\right).$ Thus, the axis of symmetry is parallel to the x-axis. It follows that:

• the vertex is $\left(h,k\right)=\left(-3,1\right)$
• the axis of symmetry is $y=k=1$
• $\mathrm{-16}=4p,$ so $p=\mathrm{-4.}$ Since $p<0,$ the parabola opens left.
• the coordinates of the focus are $\left(h+p,k\right)=\left(\mathrm{-3}+\left(\mathrm{-4}\right),1\right)=\left(\mathrm{-7},1\right)$
• the equation of the directrix is $x=h-p=\mathrm{-3}-\left(\mathrm{-4}\right)=1$
• the endpoints of the latus rectum are $\left(h+p,k\pm 2p\right)=\left(\mathrm{-3}+\left(\mathrm{-4}\right),1\pm 2\left(\mathrm{-4}\right)\right),$ or $\left(\mathrm{-7},\mathrm{-7}\right)$ and $\left(\mathrm{-7},9\right)$

Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 10.

Figure 10