C. Polynomial and Rational Functions · Rational Functions and Equations

Sketching Simple Rational Functions: A Complete Strategy

C2.3 C2.1 C2.2About 75 minutes

Learning goals

I can follow a step-by-step strategy to sketch a simple rational function from its equation.
I can find holes, vertical asymptotes and horizontal asymptotes by factoring and comparing degrees.
I can use a sign chart to decide where the graph is above or below the x-axis and how it behaves near each asymptote.
I can recognize that a graph may cross its horizontal asymptote, but never its vertical asymptote.

Words to know

degree
The highest power of x in a polynomial. 3x − 1 has degree 1; x² − 4 has degree 2.

simple rational function
In this course: the reciprocal of a linear or quadratic function, a linear function divided by a linear function, and closely related functions such as a linear function divided by a quadratic.

odd function
A function with f(−x) = −f(x). Its graph has rotational symmetry of 180° about the origin.

point of discontinuity
An x-value where the graph has a break: a vertical asymptote or a hole.

Putting it all together

In the first five lessons you studied one type of rational function at a time. Now you will sketch any simple rational function from its equation, without technology, using one reliable strategy. Graphing technology (a graphing calculator or a free online graphing tool) is great for checking a sketch, but a good hand sketch shows that you understand why the graph looks the way it does.

Five steps. Doing them in order catches holes before you mistake them for asymptotes, and the sign chart tells you which way each branch goes.

The strategy

Factor the numerator and the denominator. If a factor cancels, the graph has a hole at that x-value (not an asymptote). Find the y-coordinate of the hole from the simplified function.
Asymptotes. Vertical: the zeros of the denominator that remain after cancelling. Horizontal: compare degrees. If the numerator's degree is less than the denominator's, the HA is y = 0. If the degrees are equal, the HA is y = (ratio of the leading coefficients).
Intercepts. x-intercepts: zeros of the numerator (after cancelling). y-intercept: substitute x = 0 (if 0 is in the domain).
Sign chart. Mark all zeros and vertical asymptotes on a number line and test one value in each interval. The sign just left and right of each asymptote tells you whether the graph goes to +∞ or −∞ there.
Sketch. Draw the asymptotes as dashed lines, plot the intercepts and a few extra points (for the reciprocal of a quadratic, use the turning point), and join each branch smoothly.

(If the numerator's degree is greater than the denominator's, there is no horizontal asymptote. Those functions are not part of this course; you may meet them in university calculus.)

Worked example

Example 1: The reciprocal of a quadratic, start to finish

Sketch y = 2/(x² − x − 6).

Factor: x² − x − 6 = (x − 3)(x + 2). No common factor.
Asymptotes: VAs x = 3 and x = −2. Degree 0 < degree 2, so HA y = 0.
Intercepts: no x-intercepts (the numerator is 2). y-intercept 2/(−6) = −1/3.
Sign chart: critical values −2 and 3. x = −3 gives 2/6 > 0; x = 0 gives −1/3 < 0; x = 4 gives 2/6 > 0. So + | − | +.
Behaviour: positive just left of −2 and just right of 3, so y → +∞ there; negative just inside both asymptotes, so y → −∞ there.
Turning point: the parabola's axis is x = 0.5, and (0.5)² − 0.5 − 6 = −6.25, so the middle branch has a local maximum at (0.5, 2/(−6.25)) = (0.5, −0.32).

Three branches: positive outer branches approaching y = 0, and a middle branch below the axis with local maximum (0.5, −0.32) and y-intercept −1/3.

y = 2/((x − 3)(x + 2)). The signs + − + from the sign chart match the three branches. The local maximum (0.5, −0.32) sits on the axis of symmetry x = 0.5.

Worked example

Example 2: A hole

Sketch y = (x + 1)/(x² − 1).

Factor: (x + 1)/((x − 1)(x + 1)). The factor x + 1 cancels: y = 1/(x − 1) for x ≠ −1.
Hole at x = −1. Its y-coordinate comes from the simplified form: 1/(−1 − 1) = −1/2. Hole at (−1, −1/2).
Asymptotes: VA x = 1 only; HA y = 0.
Intercepts: no x-intercept; y-intercept 1/(0 − 1) = −1.
Sketch y = 1/(x − 1) and draw an open circle at (−1, −1/2).

The graph of y = 1/(x − 1) with an open circle at (−1, −1/2). Domain {x ∈ ℝ | x ≠ −1, 1}.

Worked example

Example 3: Equal degrees

Sketch y = 4x/(2x − 1).

Asymptotes: VA x = 1/2; degrees are equal, so HA y = 4/2 = 2.
Intercepts: both are at (0, 0).
Sign chart: critical values 0 and 1/2. x = −1: (−4)/(−3) > 0; x = 0.25: 1/(−0.5) < 0; x = 1: 4/1 > 0.
Near x = 1/2: negative on the left, so y → −∞ as x → 1/2⁻; positive on the right, so y → +∞ as x → 1/2⁺.
Extra point: x = 1 gives 4.

A hyperbola with asymptotes x = 1/2 and y = 2, passing through (0, 0) and (1, 4).

Left: y = (x + 1)/(x² − 1) is y = 1/(x − 1) with a hole at (−1, −0.5). Right: y = 4x/(2x − 1) passes through the origin and has asymptotes x = 0.5 and y = 2.

Worked example

Example 4: A graph that crosses its horizontal asymptote

Sketch y = x/(x² − 1).

Factor: x/((x − 1)(x + 1)). No common factor.
Asymptotes: VAs x = −1 and x = 1. Degree 1 < degree 2, so HA y = 0.
Intercepts: x-intercept 0 and y-intercept 0. The graph passes through its horizontal asymptote at the origin!
Sign chart: critical values −1, 0, 1. x = −2: −2/3 < 0; x = −0.5: (−0.5)/(−0.75) > 0; x = 0.5: 0.5/(−0.75) < 0; x = 2: 2/3 > 0. So − | + | − | +.
Behaviour: x → −1⁻: y → −∞; x → −1⁺: y → +∞; x → 1⁻: y → −∞; x → 1⁺: y → +∞.
Symmetry: f(−x) = −x/(x² − 1) = −f(x), so the function is odd.

Three branches. The middle one falls from +∞ (near x = −1) through the origin to −∞ (near x = 1). A horizontal asymptote describes the ends of the graph, so the graph is allowed to cross it in the middle. It can never cross a vertical asymptote, because the function is undefined there.

y = x/(x² − 1). The middle branch crosses the horizontal asymptote y = 0 at the origin. The outer branches approach y = 0 from below (left) and from above (right).

Just for fun

Sketching rational functions may be the only art class where the dashed lines are the most important part of the picture, and you are never allowed to draw on top of the vertical ones.

Hands-on activity

Sketch first, then check

You need: graph paper, a pencil, a ruler, and a graphing calculator or a free online graphing tool on a computer or phone.

Sketch these by hand using the five-step strategy: (a) y = 3/(x² − 9) (b) y = (x − 2)/(x² − 4) (c) y = (3x + 3)/(x − 1) (d) y = 1/(x² + 2).
For each one, write the asymptotes, intercepts, any hole, and the signs on your sign chart.
Now graph each with technology. Give yourself a point for each feature you had right: asymptotes, intercepts, hole, signs and shape.
For any feature you missed, write one sentence explaining what went wrong.

What to look for: (a) VAs x = ±3, HA y = 0, y-intercept −1/3, signs + − +, local max (0, −1/3). (b) Hole at (2, 1/4); VA x = −2; HA y = 0; y-intercept 1/2. (c) VA x = 1, HA y = 3, intercepts (−1, 0) and (0, −3). (d) No VAs, HA y = 0, maximum (0, 1/2). Most graphing tools do not show holes, so you must find them with algebra.

Your turn

For y = 6/(x² − 2x − 3), find the vertical and horizontal asymptotes.

For y = 6/(x² − 2x − 3), find the y-intercept and the turning point.

Describe the behaviour of y = 6/(x² − 2x − 3) on both sides of x = −1.

Find any holes and asymptotes of y = (x − 3)/(x² − 9).

Find the asymptotes, intercepts and signs of y = (2x − 4)/(x + 1).

Describe the graph of y = 1/(x² + 2x + 2).

Find the horizontal asymptote and the intercepts of y = (5x − 2)/(x² + 1).

More practice: warm-up

State the horizontal asymptote of y = 7/(x² + 3x).

State the horizontal asymptote of y = (6x + 1)/(3x − 2).

Does y = (x − 5)/(x² − 25) have a hole or a vertical asymptote at x = 5?

State the vertical asymptotes of y = 1/(x² − 3x).

Find the y-intercept of y = (x + 4)/(x − 2).

More practice: on your own

Sketch-describe y = −3/(x² − 4x): asymptotes, intercepts, signs, turning point and range.

Find the hole, asymptotes and y-intercept of y = (x + 2)/(x² + 5x + 6).

Find the asymptotes, intercepts and signs of y = (1 − 2x)/(x + 3).

Write an equation for a rational function with VAs x = −2 and x = 2, HA y = 0, no x-intercepts, y-intercept 1/2, and positive between the asymptotes.

Describe the graph of y = (x² − 1)/(x − 1). Does it have a vertical asymptote?

Challenge

Sketch-describe y = x/(x² − 9): asymptotes, intercepts, signs and symmetry.

Find a rational function with only one vertical asymptote x = 1, horizontal asymptote y = 0, a hole at x = −2, and y-intercept 3.

Find the range of y = 4x/(x² + 4). (Hint: for which y does the equation y = 4x/(x² + 4) have a real solution x?)

Check your understanding

Quiz

What is the horizontal asymptote of y = (3x − 1)/(x² + 4)?

What happens on the graph of y = (x − 4)/(x² − 16) at x = 4?

What is the y-intercept of y = −3/(x² − 4x)?

Which function has a vertical asymptote at x = 2 where y → +∞ on both sides?

Which statement about y = x/(x² − 1) is true?

Why should you factor before naming the vertical asymptotes?

For parents and teachers

Big idea (C2.3): any simple rational function can be sketched by hand with five steps: factor (watch for holes), asymptotes, intercepts, sign chart, sketch. Technology is for checking.

How to help: Ask your teen to talk you through the five steps for one function, out loud, before drawing anything. Then compare their sketch with a free online graphing tool. Note that most graphing tools do not draw holes, so a hole found by algebra is a sign of real understanding.

Common difficulties:

Treating a cancelled factor as a vertical asymptote.
Believing a graph can never cross a horizontal asymptote. It can (Example 4); it can never cross a vertical one.
Leaving the vertical asymptotes off the sign chart, which gives wrong signs.
Writing a y-intercept when x = 0 is not in the domain.