Chapter 9.3, Problem 29WE

To determine

To graph: Plot the graph for the given equation.

Explanation of Solution

Given information: It is given that $y+4\ge {\left(x+2\right)}^2$ .

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 $x=h$

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

Hence, $V\left(-2,-4\right)$

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

Hence, the parabola opens up.

Now, check the point $\left(-2,0\right)$ whether shading lies inside or outside the parabola.

  $\begin{array}{l}\Rightarrow 0+4={\left(-2+0\right)}^2\\ \Rightarrow 4=4\end{array}$

Therefore, the inside area of the curve is shaded.

Graph: The given equation $\left[y+4\ge {\left(x+2\right)}^2\right]$ is plotted below.