Chapter 9.3, Problem 27WE

To determine

To graph: Plot the graph for the given equation.

Explanation of Solution

Given information: It is given that $y<{\left(x-3\right)}^2$ .

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

Hence, $V\left(3,0\right)$

Now, $\frac{1}{4c}=1>0$

  $\therefore c=\frac{1}{4}$

The parabola opens up.

Now, check the point $\left(3,1\right)$ whether shading lies inside or outside the parabola.

  $\because 1\cancel{<}{\left(3-3\right)}^2$

Therefore, the outside area of the curve is shaded.

Graph: The given equation $\left[y<{\left(x-3\right)}^2\right]$ is plotted below.