Graphing Rational Functions Practice

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.

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.

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.

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.