Graphs of Rational Functions: Practice

The domain of a rational function drops exactly the inputs that make the denominator zero, because division by zero has no value.

$x - 7 = 0 \;\Longrightarrow\; x = 7$

Every other real number gives an honest output, so the domain is all real $x$ except $x = 7$.

Find the zero of the denominator first.

$x + 4 = 0 \;\Longrightarrow\; x = -4$

Now check whether the factor $(x+4)$ cancels. The numerator is the constant $3$, which has no factor of $(x+4)$, so nothing cancels and the factor stays in the denominator. The values therefore blow up near $x=-4$, giving a vertical asymptote at $x = -4$.

A fraction equals zero exactly when its numerator is zero and its denominator is not, so set the numerator to zero.

$x - 4 = 0 \;\Longrightarrow\; x = 4$

At $x=4$ the denominator is $5$, not $0$, so $x=4$ is in the domain and the $x$-intercept is $(4,0)$. The value $x=-1$ is the vertical asymptote, not an intercept.

Factor the denominator and set each factor to zero.

$x^2 - 9 = (x-3)(x+3) = 0 \;\Longrightarrow\; x = 3 \text{ or } x = -3$

Both values are excluded. The zero of the numerator, $x=-2$, is perfectly legal: it is where the graph crosses the $x$-axis.

The denominator is zero when $x-2 = 0$ or $x+5 = 0$, so the excluded inputs are $x=2$ and $x=-5$.

$(x-2)(x+5) = 0 \;\Longrightarrow\; x = 2 \text{ or } x = -5$

Neither factor appears in the numerator $x+1$, so neither cancels, and both excluded inputs give vertical asymptotes. The value $x=-1$ is the $x$-intercept, not an asymptote.

The degrees tie at $2$, so divide the top and the bottom by $x^2$.

$\frac{6x^2+5}{3x^2-x} = \frac{6 + \dfrac{5}{x^2}}{\,3 - \dfrac{1}{x}\,}$

Far out, the fractions with $x$ underneath them die away and what is left is the ratio of the leading coefficients.

$\frac{6}{3} = 2$

The horizontal asymptote is $y = 2$.

The denominator is zero at $x=2$, so that input is out of the domain. Factor the numerator and cancel.

$f(x) = \frac{(x-2)(x+2)}{x-2} = x+2 \qquad (x \neq 2)$

The factor $(x-2)$ cancels completely, so the graph is the line $y = x+2$ with one point missing. The missing point sits at the height the line reaches there, $2 + 2 = 4$, so the hole is at $(2,4)$.

The numerator is stuck at $5$ while the denominator grows without limit, so the fraction is a fixed amount split into more and more pieces.

$f(1000) = \frac{5}{998} \approx 0.005, \qquad f(1{,}000{,}000) \approx 0.000005$

The values shrink toward $0$ without ever reaching it, so $y=0$ is the horizontal asymptote. The degree comparison says the same thing: $0 < 1$, so the asymptote is $y=0$.

The original denominator is zero at $x=-3$, so that input is out of the domain no matter what happens next. Now count copies of $(x+3)$: one on top, one on the bottom, so it cancels completely.

$f(x) = \frac{1}{x-1} \qquad (x \neq -3,\; x \neq 1)$

Nothing is left in the denominator to make the values blow up, so the graph is the ordinary curve $y = \frac{1}{x-1}$ with the single point at $x=-3$ removed. That point sits at height $\frac{1}{-3-1} = -\frac14$, so the hole is at $\left(-3,-\frac14\right)$.

Compare the degrees before anything else: the numerator has degree $2$ and the denominator degree $1$, so the top outgrows the bottom.

$\frac{x^2+2x}{x+1} = \frac{x + 2}{\,1 + \dfrac{1}{x}\,}$

Dividing top and bottom by $x$ leaves an $x$ upstairs, so the values grow without bound and no horizontal line can catch them. (This graph does have an oblique asymptote, but that is a slanted line, not a horizontal one.)

The denominator is zero at $x=6$, so $x=6$ is out of the domain. Everywhere else the fraction is a nonzero number divided by itself.

$f(x) = \frac{x-6}{x-6} = 1 \qquad (x \neq 6)$

The factor cancels completely, so the graph is the horizontal line $y=1$ with the single point $(6,1)$ punched out. Cancelling preserved the value but did not restore the lost input.

First make sure $0$ is in the domain. The denominator factors as $(x-2)(x+1)$, whose zeros are $2$ and $-1$, so $x=0$ is allowed. Now evaluate.

$f(0) = \frac{2(0)-1}{0^2-0-2} = \frac{-1}{-2} = \frac{1}{2}$

Two negatives make a positive, so the $y$-intercept is $\left(0,\frac 12\right)$.