Graphing Rational Functions Practice

1. Let $f(x)=\dfrac{2x^2-18}{g(x)}$, where $g(x)$ is a polynomial.

   Which of the following is a possible graph of $y=f(x)$?

   Dashed lines indicate asymptotes.

   1. 
   2. 
   3. 

   4. 

2. Let $f(x)=\dfrac{ax^n+bx^2+10}{cx^m+dx-2}$, where $m$ and $n$ are integers and $a$, $b$, $c$ and $d$ are unknown constants.

   Which of the following is a possible graph of $y=f(x)$?

   Dashed lines indicate asymptotes.

   1. 
   2. 
   3. 

   4. 

3. Let $f(x)=\dfrac{12x^5-b}{6x^5+cx-d}$, where $b$, $c$, and $d$ are unknown constants.

   Which of the following is a possible graph of $y=f(x)$?

   Dashed lines indicate asymptotes.

   1. 
   2. 
   3. 

   4. 

4. Let $f(x)=\dfrac{q(x)}{x^2+12x+36}$, where $q(x)$ is a polynomial.

   Which of the following is a possible graph of $y=f(x)$?

   Dashed lines indicate asymptotes.

   1. 
   2. 
   3. 
   4. 
      

Source: Khan Academy, https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:rational-functions/x9e81a4f98389efdf:graphs-of-rational-functions/e/graphs-of-rational-functions
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1. The strategy

   Since the denominator, $g(x)$, is not given explicitly, we cannot determine anything about the discontinuities or end behavior of $f$.

   However, since we are given the numerator, we can find its zeros. On the graph, these points will have either a zero, a vertical asymptote, or a removable discontinuity.

   Determining the zeros of $f$

   The numerator of $f$ is $(2x^2-18)$, which has roots at
   $x=-3$ and $x=3$.

   Therefore, the graph of
   

   $f$ has zeros, removable discontinuities, or vertical asymptotes at $x=-3$ and $x=3$.

   Eliminating options based on the zeros of $f$

   Only Graph A has a vertical asymptote at $x=-3$ and a zero at $x=3$. This matches our criteria. None of the other graphs have zeros or discontinuities at both $x=-3$ and $x=3$, so they cannot be the graphs of $f$.

   Therefore, Graph A is the only possible graph for $f$.

   The only graph that can be a possible graph of $f$ is Graph A.

   
   

   ---

2. The strategy

   Since neither the order of the numerator nor of the denominator are given, we cannot determine anything about the end behavior of $f$.

   We also do not know anything about the roots of either the numerator or denominator.

   However, we are able to find the value of $f(0)$ by setting $x=0$. This gives us the $y$-intercept.

   Determining the $y$-intercept of $f$

   Setting $x=0$, we see that all the terms in $f$ that contain $x$ to any order become zero. In this case,  $f(0)=\dfrac{10}{-2}=-5$.

   Therefore, the $y$-intercept of $f$ is $-5$.

   Eliminating options based on the $y$-intercept of $f$
   

   Only Graph B has a $y$-intercept of $-5$. This matches our criteria. None of the other graphs have the same $y$-intercept as $f$, so they cannot be the graphs of $f$.

   Therefore, Graph B is the only possible graph for $f$.

   The only graph that can be a possible graph of $f$ is Graph B.

   

   ---

3. The strategy

   Since the values of $b$, $c$, and $d$ are not given explicitly, we cannot determine anything about the zeros or discontinuities of $f$.

   However, since we are given the coefficients of the largest-order terms in both the numerator and denominator, we can find its end behavior. Using this, we can eliminate all the incorrect options.

   Determining the end behavior of $f$

   Let's find out the end behavior of $f$ to see if it diverges, or if it has any horizontal asymptotes. This will help us to match the function with one or more of the graphs.

   $f(x)=\dfrac{12x^5-b}{6x^5+cx-d}\approx\dfrac{12x^5}{6x^5}$

   The value of $\dfrac{12x^5}{6x^5}$ does not depend on
   $x$, for $x \neq0$:

   $\dfrac{12x^5}{6x^5} =\dfrac{2\cdot\not{6}\cdot\not{x^5}}{\not{6}\cdot\not{x^5}}=2$

   Therefore, we see that $f$ has a horizontal asymptote at $y=2$.

   Eliminating options based on end behavior

   Only Graph B has a horizontal asymptote at $y=2$, so it is our only candidate.

   The only graph that can be a possible graph of $f$ is Graph B.

   

   ---

4. The strategy

   Since the numerator, $q(x)$, is not given explicitly, we cannot determine anything about the zeros or end behavior of $f$.

   However, since we are given the denominator, we can find its discontinuities. On the graph, these points will have either a vertical asymptote or a removable discontinuity.
   

   Determining the discontinuities of $f$

   The denominator of $f$ is $(x^2+12x+36)$, which has a double root at
   $x=-6$.

   Therefore, the graph of $f$ has a vertical asymptote or removable discontinuity at $x=-6$, and nowhere else.

   Eliminating options based on discontinuities
   

   Only Graph D has a vertical asymptote at $x=-6$ and nowhere else. This matches our criteria. 

   None of the other graphs have only the same discontinuity as $f$, so they cannot be the graphs of
   $f$.

   Therefore, Graph D is the only possible graph of $f$.

   The only graph that can be a possible graph of $f$ is Graph D.