Chapter 10, Problem 101RE

To determine

To shade: the given solution set

Answer to Problem 101RE

  

Explanation of Solution

Given:

  $x-y^2<4$

  

Calculation:

Consider the system of equations,

  $\{\begin{array}{l}y\ge x^2-3x\hfill \\ y\le \frac{1}{3}x-1\hfill \end{array}$

Since the system of inequalities includes the equal signs.

So, draw solid lines that indicate that the points on the lines are included in the solution set.

The Graph of $y=x^2-3x$ is a solid line as shown by red color,

The Graph of $y=\frac{1}{3}x-1$ is a solid line as shown by blue color.

  

Now choose a test point to see which side of the lines satisfies the inequality,

Test the ordered pair $(0,0)$ for $y\ge x^2-3x$ as,

  $0\ge {(0)}^2-3(0)$

  $0\ge 0$ TRUE

That means $(0,0)$ belong to the solution set of $y\ge x^2-3x$

So shade the portion of plane towards $(0,0)$ i.e. above the line $y=x^2-3x$

  

Test the ordered pair (0,0) for $y\le \frac{1}{3}x-1$ as

  $0\le \frac{1}{3}(0)-1$

  $0\le 0$ TRUE

That means $(0,0)$ belong to the solution set of $y\le \frac{1}{3}x-1$

So shade the portion of plane towards $(0,0)$ i.e. below the line $y=\frac{1}{3}x-1$

  

Therefore the graph of the solution set of the system of inequalities is given by the intersection of the graphs of the solution sets of $\{\begin{array}{l}y\ge x^2-3x\hfill \\ y\le \frac{1}{3}x-1\hfill \end{array}$

  

Conclusion:

Therefore, the graph is shaded.