Graphing Rational Functions, precalculus lesson

A graphing calculator draws a rational function by plotting many points and joining them. Near a vertical asymptote it may connect a huge positive value to a huge negative one with a steep segment that is not part of the graph, and it usually cannot show a hole at all. A sketch built from the function’s features has neither problem. Most of this page follows one function,

f(x) = \frac{2x^2 + 4x - 6}{x^2 - x - 2}.

Key terms

Domain and holes

Definition Let $r = p/q$ , where $p$ and $q$ are polynomials and $q$ is not the zero polynomial. The domain of $r$ is the set of real numbers $x$ with $q(x) \neq 0$ , and it stays the same after any simplification. Factor $p$ and $q$ and cancel every common factor to get the reduced form $P/Q$ , in which $P$ and $Q$ share no factor of positive degree. If $a$ is excluded from the domain of $r$ but $Q(a) \neq 0$ , the graph has a hole at the point

\left(a, \frac{P(a)}{Q(a)}\right).

If $Q(a) = 0$ , the line $x = a$ is a vertical asymptote.

In plain words Find where the original denominator is zero before you cancel anything. A zero that cancels away leaves one missing point in the graph, and a zero that survives cancellation gives a vertical asymptote.

Example The running function factors as

f(x) = \frac{2(x - 1)(x + 3)}{(x - 2)(x + 1)}.

Top and bottom share no factor, so this is already the reduced form. The domain is every real number except $-1$ and $2$ . There are no holes, and the lines $x = -1$ and $x = 2$ are vertical asymptotes.

Intercepts

Definition The $y$ -intercept of the graph of $r$ is the point $(0, r(0))$ , which exists only when $0$ is in the domain. The $x$ -intercepts are the points $(a, 0)$ with $a$ in the domain and $r(a) = 0$ . In terms of the reduced form $P/Q$ , these are exactly the real zeros of $P$ that lie in the domain of $r$ . A number where the original numerator and denominator are both zero is never an $x$ -intercept, because it is not in the domain. It gives a hole when $Q$ is not zero there and a vertical asymptote when $Q$ is zero there. At an $x$ -intercept the graph crosses the axis when the zero has odd multiplicity in $P$ and touches the axis without crossing when the multiplicity is even.

In plain words Put $x = 0$ into the function to see where the graph meets the $y$ -axis. Set the reduced numerator equal to zero to see where it meets the $x$ -axis, and throw out any value that is not in the domain.

Example At zero the function gives $f(0) = -6/(-2) = 3$ , so the $y$ -intercept is $(0, 3)$ . The numerator $2(x - 1)(x + 3)$ is zero at $x = 1$ and at $x = -3$ , both in the domain, so the $x$ -intercepts are $(1, 0)$ and $(-3, 0)$ . Each zero has multiplicity $1$ , so the graph crosses the axis at both.

Test intervals

Definition Mark on the number line the real zeros of $P$ and of $Q$ from the reduced form. These cut points split the line into open intervals, called test intervals. On each test interval the reduced form $P/Q$ is continuous and never zero, so it keeps one sign there. Its sign at any single test value in the interval is its sign on the whole interval, and $r$ has that same sign wherever it is defined in the interval.

In plain words A rational function can switch sign only at a zero or at a vertical asymptote. Between those places one test value tells you whether the graph runs above or below the $x$ -axis.

Example For $f$ the cut points are $-3$ , $-1$ , $1$ and $2$ .

Interval	Test value	Value of $f$	Graph runs
$(-\infty, -3)$	$-4$	$5/9$	above the axis
$(-3, -1)$	$-2$	$-3/2$	below the axis
$(-1, 1)$	$0$	$3$	above the axis
$(1, 2)$	$3/2$	$-18/5$	below the axis
$(2, \infty)$	$3$	$6$	above the axis

Every cut point here has multiplicity $1$ , so the sign flips at each one.

Approaching a vertical asymptote

Definition Let $x = a$ be a vertical asymptote, so that $Q(a) = 0$ and $P(a) \neq 0$ in the reduced form. As $x$ approaches $a$ from either side, $\lvert r(x) \rvert$ grows without bound, and the sign of $r$ on the neighboring test interval decides the direction. If $r$ is positive just to the right of $a$ , then $r(x) \to \infty$ as $x \to a^+$ , and if $r$ is negative there, then $r(x) \to -\infty$ as $x \to a^+$ . The left side works the same way with $x \to a^-$ . When $a$ is a zero of $Q$ of odd multiplicity the two sides go in opposite directions, and when the multiplicity is even they go the same way.

In plain words Next to a vertical asymptote the graph shoots up or down along the line. The test intervals already tell you which way, so no new work is needed.

Example The values of $f$ are negative on $(-3, -1)$ and positive on $(-1, 1)$ , so

\begin{gathered} f(x) \to -\infty \text{ as } x \to -1^-, \\[0.4em] f(x) \to \infty \text{ as } x \to -1^+. \end{gathered}

Around $x = 2$ the values are negative on $(1, 2)$ and positive on $(2, \infty)$ , so the graph falls toward $-\infty$ on the left of the line and comes down from $\infty$ on the right. A quick check agrees, since $f(1.9) \approx -30.4$ and $f(2.1) \approx 36.2$ .

Approaching a horizontal or slant asymptote

Definition Let $y = \ell(x)$ be the horizontal asymptote $y = L$ or the slant asymptote $y = mx + b$ of the graph of $r = P/Q$ . Subtracting $L$ in the horizontal case, or dividing $P$ by $Q$ in the slant case, shows that for every $x$ in the domain of $r$

r(x) - \ell(x) = \frac{R(x)}{Q(x)},

where $R$ is a polynomial with $\deg R < \deg Q$ . When $R$ is not the zero polynomial, this fraction keeps one sign for all large positive $x$ and one sign for all large negative $x$ . The graph approaches the asymptote from above as $x \to \infty$ if the sign for large positive $x$ is positive, and from below if it is negative. The sign for large negative $x$ decides the side as $x \to -\infty$ in the same way.

In plain words Look at the leftover fraction between the function and its asymptote. Its sign far out tells you on which side of the line the graph runs, and the two ends can differ.

Example The degrees of the numerator and denominator of $f$ are equal and the leading coefficients are $2$ and $1$ , so $y = 2$ is the horizontal asymptote. Then

\begin{aligned} &f(x) - 2 \\[0.4em] &\enspace = \frac{2x^2 + 4x - 6 - 2(x^2 - x - 2)}{x^2 - x - 2} \\[0.8em] &\enspace = \frac{6x - 2}{x^2 - x - 2}. \end{aligned}

For large positive $x$ the top and bottom are both positive, so the graph lies above $y = 2$ , and indeed $f(10) = 117/44 \approx 2.66$ . For large negative $x$ the top is negative and the bottom positive, so the graph lies below the line, and $f(-10) = 77/54 \approx 1.43$ .

Crossing a horizontal asymptote

Definition A horizontal asymptote describes the graph only as $x \to \infty$ or $x \to -\infty$ , so the graph may cross it at moderate values of $x$ . Let $r = P/Q$ be in reduced form and not constant, with horizontal asymptote $y = L$ . The graph meets the line $y = L$ exactly at the points $(a, L)$ where $a$ is in the domain of $r$ and

P(a) - L\,Q(a) = 0.

The polynomial $P - LQ$ is not the zero polynomial and has degree less than $\deg Q$ , so there are at most $\deg Q - 1$ such points. Near such an $a$ the difference $r(x) - L$ has the sign of $P(x) - L\,Q(x)$ times the fixed sign of $Q(a)$ , so the graph crosses the line at $(a, L)$ when $a$ is a zero of odd multiplicity of $P - LQ$ , and it touches the line without crossing when the multiplicity is even. The graph meets a slant asymptote in the same way, at the real zeros of the remainder $R$ that lie in the domain. A vertical asymptote $x = a$ is never crossed by the graph of a rational function, because $a$ is not in its domain.

In plain words Set the function equal to the height of the asymptote and solve. Each solution in the domain is a point where the graph meets the line, even though far out the graph only approaches it.

Example For $x$ in the domain of $f$ , the equation $f(x) = 2$ means

2x^2 + 4x - 6 = 2(x^2 - x - 2).

The $x^2$ terms cancel and $6x = 2$ , so $x = 1/3$ , which is in the domain. The graph crosses $y = 2$ at the point $(1/3, 2)$ . This matches the previous section, where the numerator $6x - 2$ of $f(x) - 2$ changes sign exactly at $x = 1/3$ . On $(-1, 1)$ the graph comes down from $\infty$ , passes through $(0, 3)$ , crosses the asymptote at $(1/3, 2)$ and reaches the $x$ -axis at $(1, 0)$ .

Putting the graph together

Definition A sketch of a rational function $r = p/q$ is complete when it shows

the domain, with every hole drawn as an open dot,
the intercepts,
the sign of $r$ on each test interval,
each vertical asymptote, with the direction of the graph on both sides of it,
the horizontal or slant asymptote, the side from which the graph approaches it at each end, and any points where the graph crosses it.

On each interval of the domain the graph is then drawn as one unbroken smooth curve that agrees with all of these facts, with a few extra plotted points wherever the shape is unclear.

In plain words Each step adds one fact about the graph. Together they fix the overall shape, so the drawing becomes a matter of connecting what you already know.

Example Take

\begin{aligned} h(x) &= \frac{x^3 - x}{x^2 - 2x - 3} \\[0.8em] &= \frac{x(x - 1)(x + 1)}{(x - 3)(x + 1)}. \end{aligned}

The domain excludes $-1$ and $3$ . The factor $x + 1$ cancels, leaving the reduced form $x(x - 1)/(x - 3)$ , so there is a hole at $x = -1$ with height $(-1)(-2)/(-4) = -1/2$ , and $x = 3$ is a vertical asymptote. The graph passes through the origin, which is the $y$ -intercept, and the $x$ -intercepts are $(0, 0)$ and $(1, 0)$ .

The cut points $0$ , $1$ and $3$ give the signs. The test values $h(-2) = -6/5$ , $h(1/2) = 1/10$ , $h(2) = -2$ and $h(4) = 12$ show that $h$ is negative on $(-\infty, 0)$ , positive on $(0, 1)$ , negative on $(1, 3)$ and positive on $(3, \infty)$ . So $h(x) \to -\infty$ as $x \to 3^-$ and $h(x) \to \infty$ as $x \to 3^+$ .

Division gives $x^2 - x = (x - 3)(x + 2) + 6$ , so for $x$ in the domain

h(x) = x + 2 + \frac{6}{x - 3}.

The slant asymptote is $y = x + 2$ . The leftover fraction is positive for $x > 3$ and negative for $x < 3$ , so the graph runs above the line on the far right and below it on the far left. The fraction is never zero, so the graph never crosses the slant asymptote.

Putting it together, the graph comes in from the lower left just under the line $y = x + 2$ , rises through the open dot at $(-1, -1/2)$ to the origin, makes a small hump above the axis between $0$ and $1$ , and drops through $(1, 0)$ down along $x = 3$ . To the right of $x = 3$ it comes down from $\infty$ , turns, and then follows $y = x + 2$ from above.

Check yourself

Where does the graph of the function below cross its horizontal asymptote?

r(x) = \frac{3x^2 - x}{x^2 + 1}

Show the answer

The degrees are equal and the leading coefficients are $3$ and $1$ , so the horizontal asymptote is $y = 3$ . Subtracting it gives

\begin{aligned} r(x) - 3 &= \frac{3x^2 - x - 3(x^2 + 1)}{x^2 + 1} \\[0.8em] &= \frac{-x - 3}{x^2 + 1}. \end{aligned}

The denominator is never zero, and the numerator $-x - 3$ is zero only at $x = -3$ , where it changes sign. So the graph crosses its asymptote exactly once, at $(-3, 3)$ .

Describe the graph of the function below near $x = 1$ and far out to the left and to the right.

r(x) = \frac{x + 2}{(x - 1)^2}

Show the answer

The denominator is positive for every $x \neq 1$ and the numerator is close to $3$ near $x = 1$ , so $r(x) \to \infty$ from both sides of the vertical asymptote $x = 1$ . The degree of the numerator is smaller, so $y = 0$ is the horizontal asymptote. The function has the sign of $x + 2$ , so the graph approaches $y = 0$ from above as $x \to \infty$ and from below as $x \to -\infty$ . It crosses this asymptote once, at the $x$ -intercept $(-2, 0)$ .

Find the holes, intercepts and asymptotes of the function below.

r(x) = \frac{x^2 + 2x}{x^2 - 4}

Show the answer

Factoring gives

r(x) = \frac{x(x + 2)}{(x - 2)(x + 2)},

so the domain excludes $-2$ and $2$ . The reduced form is $x/(x - 2)$ . It equals $-2/(-4) = 1/2$ at $-2$ , so there is a hole at $(-2, 1/2)$ , and $x = 2$ is the vertical asymptote. The only intercept is the origin. The horizontal asymptote is $y = 1$ , and the graph never crosses it, since $x = x - 2$ has no solution.

Graphing Rational Functions

Key terms

Domain and holes

Intercepts

Test intervals

Approaching a vertical asymptote

Approaching a horizontal or slant asymptote

Crossing a horizontal asymptote

Putting the graph together

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus