Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
2. Graphing Rational Functions
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Exercise 46 Page 371

Practice makes perfect
a To graph the function, we should first determine its asymptotes. The function is in the form of y= ax+bcx+d where x=- dc is the vertical asymptote and y= ac is the horizontal asymptote.
y=15x/1.1-x ⇔ y=15x+ 0/-1x+ 1.1 Therefore, it has a vertical asymptote at x=1.1 and a horizontal asymptote at y=-15. Next, we will make a table of values.
x 15x/1.1-x y
-8 15( -8)/1.1-( -8) -13.19
-4 15( -4)/1.1-( -4) -11.76
0 15( 0)/1.1- 0 0
1.1 15( 1.1)/1.1- 1.1 Asymptote
2 15( 2)/1.1- 2 -33.33
6 15( 6)/1.1- 6 -18.37
10 15( 10)/1.1- 10 -16.85

Next, we will plot the points and draw two branches of the hyperbola so that they pass through the plotted points and approach the asymptotes.

To determine the domain and range, we exclude the vertical asymptote from the domain and the horizontal asymptote from the range. Domain:& All real numbers except 1.1 Range:& All real numbers except -15

b To find the cost of removing 20 % of the pollutant, we will substitute x= 0.2 in the given function. Then we will solve the resulting equation for y.
y=15x/1.1-x
y=15( 0.2)/1.1- 0.2
Evaluate right-hand side
y=3/1.1-0.2
y=3/0.9
y=3.33333...
y≈3.33
The cost of removing 20 % of the pollutant is about $3333. Proceeding in the same way, we can also find the cost of removing 40 % and 80 % of the pollutant.
Percent x 15x/1.1-x y
20 0.2 15( 0.2)/1.1- 0.2 3.33
40 0.4 15( 0.4)/1.1- 0.4 8.57
80 0.8 15( 0.8)/1.1- 0.8 40

As we can see, doubling the percentage of the pollutant does not double the cost.