Concept explainers
a)
To write: An equation that can be solved to find the points of intersection of the graphs of y1 and y2 .
An equation that can be solved to find the points of intersection of the graphs of y1 and y2 is −x3−2x2+5x+2=0
Given Information:
The equations y1=4x+5 , y2=x3+2x2−x+3 and y3=−x3−2x2+5x+2 .
Concept Used:
The points of intersection of graphs of y1 and y2 are the values of x where y1=y2 .
Calculation:
Equate y1=4x+5 to y2=x3+2x2−x+3 and simplify.
y1=y24x+5=x3+2x2−x+3x3+2x2−x+3−4x−5=0x3+2x2−5x−2=0−x3−2x2+5x+2=0
Therefore, −x3−2x2+5x+2=0 is an equation that can be solved to find the points of intersection of the graphs of y1 and y2 .
b)
To find: Write an equation that can be solved to find the x -intercepts of the graph of y .
The required equation is −x3−2x2+5x+2=0 .
Given Information:
The equation y3=−x3−2x2+5x+2 .
Concept Used:
The intercept for any equation is found by equation it to 0.
Calculation:
Equate y3 to 0.
y3=0−x3−2x2+5x+2=0
Therefore, the required equation is −x3−2x2+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 y1=4x+5 , y2=x3+2x2−x+3 , y3=−x3−2x2+5x+2 and the graph shown in figure 1.
Figure 1
Explanation:
Draw the vertical lines at the points of intersection of y1 and y2 as shown in figure 2.
Figure 2
From figure 2 it follows that the value of x at which the graphs of y1 and y2 intersects are the x -intercepts of the graph of y3=−x3−2x2+5x+2 .
Also, the equation for finding the intersection points of y1 and y2 is same equation that can be used to find the x -intercepts y3=−x3−2x2+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 y3=−x3−2x2+5x+2 give the same values when substituted into the expressions for and y1 and y2 .
Given Information:
The equations y1=4x+5 , y2=x3+2x2−x+3 and y3=−x3−2x2+5x+2 .
Calculation:
The equation −x3−2x2+5x+2=0 gives the x -intercepts of the graph of y3 .
The graph of −x3−2x2+5x+2 is shown in figure 3.
Figure 3
Thus, the real solutions of −x3−2x2+5x+2=0 are −3.323 , −0.358 and 1.681 .
Now, substitute x=−3.323 , x=−0.358 and x=1.681 into y2=x3+2x2−x+3 as follows.
y1(−3.323)=4(−3.323)+5=−8.292y1(−0.358)=4(−0.358)+5=3.568y1(1.681)=4(1.681)+5=11.724
Now, substitute x=−3.323 , x=−0.358 and x=1.681 into y1=4x+5 as follows. y2(−3.323)=(−3.323)3+2(−3.323)2−(−3.323)+3=−8.286y2(−0.358)=(−0.358)3+2(−0.358)2−(−0.358)+3=−3.568y2(1.681)=(1.681)3+2(1.681)2−(1.681)+3=11.721
Thus, y1=y2 at the points −3.323 , −0.358 and 1.681 .
Therefore, it follows that the intercepts of y3=−x3−2x2+5x+2 aret the points of intersection of y1 and y2 .
Chapter 1 Solutions
PRECALCULUS:...COMMON CORE ED.-W/ACCESS
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