Exercise 21 Page 647

First, notice that the variable raised to the power of $2$ is $x.$ Therefore, the axis of symmetry will be a vertical line. We will start by writing the given equation in the form $y=a(x-h{)}^2+k.$ To do so, we will first rearrange the equation so that the $x\text{-}$terms and the constant are on the right-hand side. We can now simplify the right-hand side by completing the square. We have to add and subtract ${\left(\frac{b}{2}\right)}^2.$ In this case, we have that the linear coefficient $b$ is $8.$ Let's do it! Now, let's identify the values of $a,$ $h,$ and $k.$ We can see that $a=\frac{1}{3},$ $h=\text{-}4,$ and that $k=\text{-}9.$ Next, we can use this information to highlight some important characteristics of the parabola.

	$y=a(x-h{)}^2+k$	$y=\frac{1}{3}(x-(\text{-}4){)}^2+(\text{-}9)$
Direction of Opening	Up if $a>0,$ Down if $a<0$	Up $\left(2>0\right)$
Vertex	$(h,k)$	$(\text{-}4,\text{-}9)$
Axis of Symmetry	$x=h$	$x=\text{-}4$
Focus	$\left(h,k+\frac{1}{4a}\right)$	$\left(\text{-}4,\text{-}9+\frac{1}{4(\frac{1}{3})}\right)$ $\Updownarrow$ $(\text{-}4,\text{-}8.25)$
Directrix	$y=k-\frac{1}{4a}$	$y=\text{-}9-\frac{1}{4(\frac{1}{3})}$ $\Updownarrow$ $y=\text{-}9.75$
Length of Latus Rectum	$\left|\frac{1}{a}\right|$ units	$\left|\frac{1}{\frac{1}{3}}\right|$ units $\Updownarrow$ $3$ units

We can use the above information to draw the graph.