Chapter 9.3, Problem 30WE

To determine

To graph: Plot the graph for the given equation.

Explanation of Solution

Given information: It is given that $x-11<y^2+6y$ .

Calculation: Since for parabola $y-k=\frac{1}{4c}{\left(x-h\right)}^2$ :

  $\begin{array}{l}V\left(h,k\right)\\ F\left(h,k+c\right)\\ D:y=k-c\end{array}$

And axis of symmetry $x=h$

Rewrite the given equation,

  $\begin{array}{l}x-11<y^2+6y\\ x-11+9<y^2+6y+9\\ x-2<{\left(y+3\right)}^2\end{array}$

Compare the given equation $x-2<{\left(y+3\right)}^2$ to $y-k=\frac{1}{4c}{\left(x-h\right)}^2$

Hence, $V\left(2,-3\right)$

Since $\frac{1}{4c}=1>0$

Hence, the parabola opens right.

Now, check the point $\left(3,-3\right)$ whether shading lies inside or outside the parabola.

  $\begin{array}{l}\because 3-11<{\left(-3\right)}^2+6\times -3\\ 9<9-18\\ 9\cancel{<}9\end{array}$

Therefore, the outside area of the curve is shaded.

Graph: The given equation $\left[x-11<y^2+6y\right]$ is plotted below.