Chapter 1.1, Problem 48E

Solution

a)

To write: An equation that can be solved to find the points of intersection of the graphs of $y_1$ and $y_2$ .

An equation that can be solved to find the points of intersection of the graphs of $y_1$ and $y_2$ is $-x^3-2x^2+5x+2=0$

Given Information:

The equations $y_1=4x+5$ , $y_2=x^3+2x^2-x+3$ and $y_3=-x^3-2x^2+5x+2$ .

Concept Used:

The points of intersection of graphs of $y_1$ and $y_2$ are the values of $x$ where $y_1=y_2$ .

Calculation:

Equate $y_1=4x+5$ to $y_2=x^3+2x^2-x+3$ and simplify.

  $\begin{array}{c}{y}_1=y_2\\ 4x+5=x^3+2x^2-x+3\\ x^3+2x^2-x+3-4x-5=0\\ x^3+2x^2-5x-2=0\\ -x^3-2x^2+5x+2=0\end{array}$

Therefore, $-x^3-2x^2+5x+2=0$ is an equation that can be solved to find the points of intersection of the graphs of $y_1$ and $y_2$ .

b)

To find: Write an equation that can be solved to find the x -intercepts of the graph of y .

The required equation is $-x^3-2x^2+5x+2=0$ .

Given Information:

The equation $y_3=-x^3-2x^2+5x+2$ .

Concept Used:

The intercept for any equation is found by equation it to 0.

Calculation:

Equate $y_3$ to 0.

  $\begin{array}{c}{y}_3=0\\ -x^3-2x^2+5x+2=0\end{array}$

Therefore, the required equation is $-x^3-2x^2+5x+2=0$ .

c)

To explain: How does the graphical model reflect the fact that the answers to (a) and (b) are equivalent algebraically.

Given Information:

The equations $y_1=4x+5$ , $y_2=x^3+2x^2-x+3$ , $y_3=-x^3-2x^2+5x+2$ and the graph shown in figure 1.

  

Figure 1

Explanation:

Draw the vertical lines at the points of intersection of $y_1$ and $y_2$ as shown in figure 2.

  

Figure 2

From figure 2 it follows that the value of $x$ at which the graphs of $y_1$ and $y_2$ intersects are the $x$ -intercepts of the graph of $y_3=-x^3-2x^2+5x+2$ .

Also, the equation for finding the intersection points of $y_1$ and $y_2$ is same equation that can be used to find the $x$ -intercepts $y_3=-x^3-2x^2+5x+2$ .

Therefore, the graphical model reflects the fact that the answers to (a) and (b) are equivalent algebraically.

d)

To check: Numerically that the $x$ -intercepts of $y_3=-x^3-2x^2+5x+2$ give the same values when substituted into the expressions for and $y_1$ and $y_2$ .

Given Information:

The equations $y_1=4x+5$ , $y_2=x^3+2x^2-x+3$ and $y_3=-x^3-2x^2+5x+2$ .

Calculation:

The equation $-x^3-2x^2+5x+2=0$ gives the x -intercepts of the graph of $y_3$ .

The graph of $-x^3-2x^2+5x+2$ is shown in figure 3.

  

Figure 3

Thus, the real solutions of $-x^3-2x^2+5x+2=0$ are $-3.323$ , $-0.358$ and $1.681$ .

Now, substitute $x=-3.323$ , $x=-0.358$ and $x=1.681$ into $y_2=x^3+2x^2-x+3$ as follows.

  $\begin{array}{c}{y}_1\left(-3.323\right)=4\left(-3.323\right)+5\\ =-8.292\\ y_1\left(-0.358\right)=4\left(-0.358\right)+5\\ =3.568\\ y_1\left(1.681\right)=4\left(1.681\right)+5\\ =11.724\end{array}$

Now, substitute $x=-3.323$ , $x=-0.358$ and $x=1.681$ into $y_1=4x+5$ as follows. $\begin{array}{c}{y}_2\left(-3.323\right)={\left(-3.323\right)}^3+2{\left(-3.323\right)}^2-\left(-3.323\right)+3\\ =-8.286\\ y_2\left(-0.358\right)={\left(-0.358\right)}^3+2{\left(-0.358\right)}^2-\left(-0.358\right)+3\\ =-3.568\\ y_2\left(1.681\right)={\left(1.681\right)}^3+2{\left(1.681\right)}^2-\left(1.681\right)+3\\ =11.721\end{array}$

Thus, $y_1=y_2$ at the points $-3.323$ , $-0.358$ and $1.681$ .

Therefore, it follows that the intercepts of $y_3=-x^3-2x^2+5x+2$ aret the points of intersection of $y_1$ and $y_2$ .