Chapter 4.9, Problem 52PPSE

Solution

a.

To Sketch: The graphs of $y=2x^2+4x-5$ and $y=x^2+2x-3$ . Change these equations into inequalities so the system has solutions that comprise two non-overlapping regions.

Given:

The graphs of $y=2x^2+4x-5$ and $y=x^2+2x-3$

Explanation:

The graphical representation of $y=2x^2+4x-5$ and $y=x^2+2x-3$ :

It appears that the two graphs can be divided into five regions based on the $x-y$ planes, and

five points have been plotted so that one point lies within each of these regions:

  

All points in a region will satisfy the inequality if a point in it satisfies it.

  $y=2x^2+4x-5$ can be changed as inequalities to:

  $y>2x^2+4x-5$ and $y<2x^2+4x-5$

Check the points which satisfy $y>2x^2+4x-5:$

Take the point $\left(-1,1\right)$

  $\begin{array}{l}1>2{\left(-1\right)}^2+4\left(-1\right)-5\\ 1>2-4-5\\ 1>-7\end{array}$

The inequality satisfied.

Take the point $\left(-1,-5\right)$

  $\begin{array}{l}-5>2{\left(-1\right)}^2+4\left(-1\right)-5\\ -5>2-4-5\\ -5>-7\end{array}$

The inequality satisfied.

Take the point $\left(-5,-5\right)$

  $\begin{array}{l}-5>2{\left(-5\right)}^2+4\left(-5\right)-5\\ -5>50-20-5\\ -5>25\end{array}$

The inequality does not satisfy.

Take the point $\left(1.7,5\right)$

  $\begin{array}{l}5>2{\left(1.7\right)}^2+4\left(1.7\right)-5\\ 5>5.78+6.8-5\\ 5>7.58\end{array}$

The inequality does not satisfy.

Take the point $\left(-3.7,5\right)$

  $\begin{array}{l}5>2{\left(-3.7\right)}^2+4\left(-3.7\right)-5\\ 5>27.38-14.8-5\\ 5>7.58\end{array}$

The inequality does not satisfy.

The points which do not satisfy the Inequality $y>2x^2+4x-5$ satisfy the Inequality $y<2x^2+4x-5$ . Because, the parabola does not contain any points.

Check for the points which satisfy $y>x^2+2x-3$ :

Take the point $\left(-1,1\right)$

  $\begin{array}{l}1>{\left(-1\right)}^2+2\left(-1\right)-3\\ 1>1-2-3\\ 1>-4\end{array}$

The inequality satisfied.

Take the point $\left(-1,-5\right)$

  $\begin{array}{l}-5>{\left(-1\right)}^2+2\left(-1\right)-3\\ -5>1-2-3\\ -5>-4\end{array}$

The inequality does not satisfy.

Take the point $\left(-5,-5\right)$

  $\begin{array}{l}-5>{\left(-5\right)}^2+2\left(-5\right)-3\\ -5>25-10-3\\ -5>12\end{array}$

The inequality does not satisfy.

Take the point $\left(1.7,5\right)$

  $\begin{array}{l}5>{\left(1.7\right)}^2+2\left(1.7\right)-3\\ 5>2.89+3.4-3\\ 5>3.3\end{array}$

The inequality satisfied.

Take the point $\left(-3.7,5\right)$

  $\begin{array}{l}5>2{\left(-3.7\right)}^2+4\left(-3.7\right)-5\\ 5>13.69-7.4-3\\ 5>3.29\end{array}$

The inequality satisfied.

The points which do not satisfy the Inequality $y>x^2+2x-3$ satisfy the Inequality $y<x^2+2x-3$ . Because, the parabola does not contain any points.

Find the Inequalities whose solutions are the points in two non-overlapping regions:

The Inequalities are:

  $y<2x^2+4x-5$ and $y>x^2+2x-3$ satisfy the regions which consists $\left(1.7,5\right)$ and $\left(-3.7,5\right)$

So, the pair of required inequalities are:

  $y<2x^2+4x-5$ And $y>x^2+2x-3$

The graphical representation is:

  

b.

To Sketch: The graphs of $y=2x^2+4x-5$ and $y=x^2+2x-3$ . Change these equations into inequalities so the system has solutions that comprise one bounded region.

Given:

The graphs of $y=2x^2+4x-5$ and $y=x^2+2x-3$

Explanation:

The graphical representation of $y=2x^2+4x-5$ and $y=x^2+2x-3$ :

It appears that the two graphs can be divided into five regions based on the $x-y$ planes, and five points have been plotted so that one point lies within each of these regions:

  

All points in a region will satisfy the inequality if a point in it satisfies it.

  $y=2x^2+4x-5$ can be changed as inequalities to:

  $y>2x^2+4x-5$ and $y<2x^2+4x-5$

Check the points which satisfy $y>2x^2+4x-5:$

Take the point $\left(-1,1\right)$

  $\begin{array}{l}1>2{\left(-1\right)}^2+4\left(-1\right)-5\\ 1>2-4-5\\ 1>-7\end{array}$

The inequality satisfied.

Take the point $\left(-1,-5\right)$

  $\begin{array}{l}-5>2{\left(-1\right)}^2+4\left(-1\right)-5\\ -5>2-4-5\\ -5>-7\end{array}$

The inequality satisfied.

Take the point $\left(-5,-5\right)$

  $\begin{array}{l}-5>2{\left(-5\right)}^2+4\left(-5\right)-5\\ -5>50-20-5\\ -5>25\end{array}$

The inequality does not satisfy.

Take the point $\left(1.7,5\right)$

  $\begin{array}{l}5>2{\left(1.7\right)}^2+4\left(1.7\right)-5\\ 5>5.78+6.8-5\\ 5>7.58\end{array}$

The inequality does not satisfy.

Take the point $\left(-3.7,5\right)$

  $\begin{array}{l}5>2{\left(-3.7\right)}^2+4\left(-3.7\right)-5\\ 5>27.38-14.8-5\\ 5>7.58\end{array}$

The inequality does not satisfy.

The points which do not satisfy the Inequality $y>2x^2+4x-5$ satisfy the Inequality $y<2x^2+4x-5$ . Because, the parabola does not contain any points.

Check for the points which satisfy $y>x^2+2x-3$ :

Take the point $\left(-1,1\right)$

  $\begin{array}{l}1>{\left(-1\right)}^2+2\left(-1\right)-3\\ 1>1-2-3\\ 1>-4\end{array}$

The inequality satisfied.

Take the point $\left(-1,-5\right)$

  $\begin{array}{l}-5>{\left(-1\right)}^2+2\left(-1\right)-3\\ -5>1-2-3\\ -5>-4\end{array}$

The inequality does not satisfy.

Take the point $\left(-5,-5\right)$

  $\begin{array}{l}-5>{\left(-5\right)}^2+2\left(-5\right)-3\\ -5>25-10-3\\ -5>12\end{array}$

The inequality does not satisfy.

Take the point $\left(1.7,5\right)$

  $\begin{array}{l}5>{\left(1.7\right)}^2+2\left(1.7\right)-3\\ 5>2.89+3.4-3\\ 5>3.3\end{array}$

The inequality satisfied.

Take the point $\left(-3.7,5\right)$

  $\begin{array}{l}5>2{\left(-3.7\right)}^2+4\left(-3.7\right)-5\\ 5>13.69-7.4-3\\ 5>3.29\end{array}$

The inequality satisfied.

The points which do not satisfy the Inequality $y>x^2+2x-3$ satisfy the Inequality $y<x^2+2x-3$ . Because, the parabola does not contain any points.

Find the Inequalities whose solutions are the points in a bounded region:

From the figure, the only Bounded Region is which has the points $\left(-1,-5\right)$

From above, the Inequalities are:

  $y>2x^2+4x-5$ and $y<x^2+2x-3$ satisfy the Regions which has the points $\left(-1,-5\right)$

So, the needed pair of inequalities are:

  $y>2x^2+4x-5$ and $y<x^2+2x-3$

The graphical representation is: