Chapter 8.1, Problem 33E

Solution

To sketch: the graph of the parabola ${\left(x+4\right)}^2=-12\left(y+1\right)$ by hand.

Given:

The equation of a parabola: ${\left(x+4\right)}^2=-12\left(y+1\right)$ .

Graph:

The general form of the equation of a parabola that has the vertex $\left(h,k\right)$ and focus $\left(h,k+p\right)$ is ${\left(x-h\right)}^2=4p\left(y-k\right)$ .

Observe that the equation ${\left(x+4\right)}^2=-12\left(y+1\right)$ is in the form ${\left(x-h\right)}^2=4p\left(y-k\right)$ where $h=-4$ , $k=-1$ and $p=-3$ .

Thus, ${\left(x+4\right)}^2=-12\left(y+1\right)$ is the equation of the parabola having the vertex $\left(-4,-1\right)$ and focus $\left(-4,-1+\left(-3\right)\right)=\left(-4,-4\right)$ .

Plot the vertex $\left(-4,-1\right)$ and the focus $\left(-4,-4\right)$ of the parabola on a coordinate grid as shown below:

  

Since the parabola has the vertex $\left(-4,-1\right)$ and the focus $\left(-4,-4\right)$ , it is opened to the lower side of the grid.

That is, the possible $y$ values of the parabola ${\left(x+4\right)}^2=-12\left(y+1\right)$ are all negative.

Now, find few points on the parabola for the purpose of drawing it accurately.

For that purpose, substitute $y=-4$ in the equation ${\left(x+4\right)}^2=-12\left(y+1\right)$ and solve it for $x$ :

  $\begin{array}{c}{\left(x+4\right)}^2=-12\left(-4+1\right)\\ {\left(x+4\right)}^2=36\\ x+4=\pm \sqrt{36}\\ x+4=6,-6\\ x=6-4,-6-4\\ x=2,-10\end{array}$ .

Thus, the points $\left(-10,-4\right)$ and $\left(2,-4\right)$ lie on the parabola ${\left(x+4\right)}^2=-12\left(y+1\right)$ .

Similarly, it may be observed that the points $\left(8,-13\right)$ and $\left(-16,-13\right)$ lie on the parabola ${\left(x+4\right)}^2=-12\left(y+1\right)$ .

Plot the points $\left(-10,-4\right)$ , $\left(2,-4\right)$ , $\left(8,-13\right)$ and $\left(-16,-13\right)$ along with the vertex and the focus of the parabola on the same grid:

  

Then, join the points plotted previously and the vertex of the parabola using a smooth curve to obtain the graph of the parabola as shown below: