Chapter 2, Problem 10PS

a.

To determine

Match the graph of the rational function.

Answer to Problem 10PS

  $\boxed{\begin{array}{l}a<0\\ b>0\\ c>0\\ d<0\end{array}}$

Explanation of Solution

Given information:

Match the graph of the rational function

  $f(x)=\frac{ax+b}{cx+d}$

with the given conditions.

  

  $\begin{array}{l}a>0\\ b<0\\ c>0\\ d<0\end{array}$

  $\begin{array}{l}a>0\\ b>0\\ c<0\\ d<0\end{array}$

  $\begin{array}{l}a<0\\ b>0\\ c>0\\ d<0\end{array}$

  $\begin{array}{l}a>0\\ b<0\\ c>0\\ d>0\end{array}$

Calculation:

Consider the rationl function

  $f\left(x\right)=\frac{ax+b}{cx+d}$

Put $x=0$ in the above function to find the y-intercept of the function

  $\begin{array}{l}f\left(0\right)=\frac{a\left(0\right)+b}{c\left(0\right)+d}\\ =\frac{b}{d}\end{array}$

Thus, the y-intercept of the function is $\left(0,\frac{b}{d}\right)$

Put $f\left(x\right)=0$ in the above function to find the x-intercept of the function

  $0=\frac{ax+b}{cx+d}$

  $x=-\frac{b}{a}$

Therefore, the x-intercept of the function is $\left(-\frac{b}{a},0\right)$

The vertical asymptotes are given by the zeroes of the denominator

Therefore

  $\begin{array}{l}cx+d=0\\ x=-\frac{d}{c}\\ \end{array}$

The horizontal asymptote is given by the ration of the leading coefficients

Thus, the horizontal asymptote is $y=\frac{a}{c}$

Consider the following graph

  

Note that, it has positive vertical asymptote and negative horizontal asymptote. Also it is having positive $x$ intercept and negative $y$ intercept.

Therefore

  $\begin{array}{l}a<0\\ b>0\\ c>0\\ d<0\end{array}$

Hence, option (iii) is the correct answer.

b.

To determine

Match the graph of the rational function.

Answer to Problem 10PS

  $\boxed{\begin{array}{l}a>0\\ b>0\\ c<0\\ d<0\end{array}}$

Explanation of Solution

Given information:

Match the graph of the rational function

  $f(x)=\frac{ax+b}{cx+d}$

with the given conditions.

  

  $\begin{array}{l}a>0\\ b<0\\ c>0\\ d<0\end{array}$

  $\begin{array}{l}a>0\\ b>0\\ c<0\\ d<0\end{array}$

  $\begin{array}{l}a<0\\ b>0\\ c>0\\ d<0\end{array}$

  $\begin{array}{l}a>0\\ b<0\\ c>0\\ d>0\end{array}$

Calculation:

Consider the rationl function

  $f\left(x\right)=\frac{ax+b}{cx+d}$

Put $x=0$ in the above function to find the y-intercept of the function

  $\begin{array}{l}f\left(0\right)=\frac{a\left(0\right)+b}{c\left(0\right)+d}\\ =\frac{b}{d}\end{array}$

Thus, the y-intercept of the function is $\left(0,\frac{b}{d}\right)$

Put $f\left(x\right)=0$ in the above function to find the x-intercept of the function

  $0=\frac{ax+b}{cx+d}$

  $x=-\frac{b}{a}$

Therefore, the x-intercept of the function is $\left(-\frac{b}{a},0\right)$

The vertical asymptotes are given by the zeroes of the denominator

Therefore

  $\begin{array}{l}cx+d=0\\ x=-\frac{d}{c}\\ \end{array}$

The horizontal asymptote is given by the ration of the leading coefficients

Thus, the horizontal asymptote is $y=\frac{a}{c}$

Consider the following graph

  

Note that it has negative vertical asymptote and negative horizontal asymptote. Also, it is having negative $x$ intercept and negative $y$ intercept.

  $\begin{array}{l}a>0\\ b>0\\ c<0\\ d<0\end{array}$

Hence, option (ii)is the correct answer.

c.

To determine

Match the graph of the rational function.

Answer to Problem 10PS

  $\boxed{\begin{array}{l}a>0\\ b<0\\ c>0\\ d>0\end{array}}$

Explanation of Solution

Given information:

Match the graph of the rational function

  $f(x)=\frac{ax+b}{cx+d}$

with the given conditions.

  

  $\begin{array}{l}a>0\\ b<0\\ c>0\\ d<0\end{array}$

  $\begin{array}{l}a>0\\ b>0\\ c<0\\ d<0\end{array}$

  $\begin{array}{l}a<0\\ b>0\\ c>0\\ d<0\end{array}$

  $\begin{array}{l}a>0\\ b<0\\ c>0\\ d>0\end{array}$

Calculation:

Consider the rationl function

  $f\left(x\right)=\frac{ax+b}{cx+d}$

Put $x=0$ in the above function to find the y-intercept of the function

  $\begin{array}{l}f\left(0\right)=\frac{a\left(0\right)+b}{c\left(0\right)+d}\\ =\frac{b}{d}\end{array}$

Thus, the y-intercept of the function is $\left(0,\frac{b}{d}\right)$

Put $f\left(x\right)=0$ in the above function to find the x-intercept of the function

  $0=\frac{ax+b}{cx+d}$

  $x=-\frac{b}{a}$

Therefore, the x-intercept of the function is $\left(-\frac{b}{a},0\right)$

The vertical asymptotes are given by the zeroes of the denominator

Therefore

  $\begin{array}{l}cx+d=0\\ x=-\frac{d}{c}\\ \end{array}$

The horizontal asymptote is given by the ration of the leading coefficients

Thus, the horizontal asymptote is $y=\frac{a}{c}$

Consider the following graph

  

Note that it has negative vertical asymptote and positive horizontal asymptote. Also, it is has negative $x$ intercept and positive $y$ intercept.

Hence,

  $\begin{array}{l}a>0\\ b<0\\ c>0\\ d>0\end{array}$

Hence, option (iv) is the correct answer.

d.

To determine

Match the graph of the rational function.

Answer to Problem 10PS

  $\boxed{\begin{array}{l}a>0\\ b<0\\ c>0\\ d<0\end{array}}$

Explanation of Solution

Given information:

Match the graph of the rational function

  $f(x)=\frac{ax+b}{cx+d}$

with the given conditions.

  

  $\begin{array}{l}a>0\\ b<0\\ c>0\\ d<0\end{array}$

  $\begin{array}{l}a>0\\ b>0\\ c<0\\ d<0\end{array}$

  $\begin{array}{l}a<0\\ b>0\\ c>0\\ d<0\end{array}$

  $\begin{array}{l}a>0\\ b<0\\ c>0\\ d>0\end{array}$

Calculation:

Consider the rationl function

  $f\left(x\right)=\frac{ax+b}{cx+d}$

Put $x=0$ in the above function to find the y-intercept of the function

  $\begin{array}{l}f\left(0\right)=\frac{a\left(0\right)+b}{c\left(0\right)+d}\\ =\frac{b}{d}\end{array}$

Thus, the y-intercept of the function is $\left(0,\frac{b}{d}\right)$

Put $f\left(x\right)=0$ in the above function to find the x-intercept of the function

  $0=\frac{ax+b}{cx+d}$

  $x=-\frac{b}{a}$

Therefore, the x-intercept of the function is $\left(-\frac{b}{a},0\right)$

The vertical asymptotes are given by the zeroes of the denominator

Therefore

  $\begin{array}{l}cx+d=0\\ x=-\frac{d}{c}\\ \end{array}$

The horizontal asymptote is given by the ration of the leading coefficients

Thus, the horizontal asymptote is $y=\frac{a}{c}$

Consider the following graph

  

Note that it has positive vertical asymptote and positive horizontal asymptote. Also, it is has positive $x$ intercept and positive $y$ intercept.

Hence,

  $\begin{array}{l}a>0\\ b<0\\ c>0\\ d<0\end{array}$

Hence, option (i) is the correct answer.