C. Polynomial and Rational Functions · Rational Functions and Equations

Rational Functions y = (ax + b)/(cx + d): Asymptotes and Intercepts

C2.2About 70 minutes

Learning goals

I can find the vertical asymptote, horizontal asymptote and intercepts of a rational function with a linear numerator and denominator.
I can explain why the horizontal asymptote is y = a/c by looking at very large values of x.
I can describe the behaviour near the vertical asymptote and decide whether the graph is above or below the horizontal asymptote at each end.
I can recognize when a common factor makes a hole instead of an asymptote.

Words to know

leading coefficient
The coefficient of the highest power of x. In 2x + 1 it is 2.

end behaviour
What the y-values do as x → +∞ and as x → −∞.

hole (point of discontinuity)
A single missing point on a graph, caused by a factor that cancels from the numerator and denominator. It is drawn as an open circle.

hyperbola
The two-branch curve shape of y = 1/x and its transformations, including every non-degenerate y = (ax + b)/(cx + d).

Batting averages and asymptotes

A baseball player has 3 hits in 10 at-bats. If she gets a hit in every one of her next x at-bats, her average becomes A(x) = (3 + x)/(10 + x). After 10 straight hits it is 13/20 = 0.65. After 90 straight hits it is 93/100 = 0.93. However many hits in a row she gets, the average creeps toward 1 (a perfect 1.000) but never reaches it, because the denominator is always 7 more than the numerator.

Notice the shape of A(x): a linear expression divided by a linear expression. These functions have a vertical asymptote and a horizontal asymptote that is not y = 0.

Finding the features of y = (ax + b)/(cx + d)

Take y = (2x + 1)/(x − 3).

Vertical asymptote: where the denominator is 0 (and the numerator is not): x − 3 = 0, so x = 3.
x-intercept: where the numerator is 0: 2x + 1 = 0, so x = −1/2.
y-intercept: put x = 0: (0 + 1)/(0 − 3) = −1/3.
Horizontal asymptote: look at very large x. Divide the numerator and denominator by x:

y = (2 + 1/x)/(1 − 3/x).

As x → ±∞, 1/x and 3/x go to 0, so y → 2/1 = 2. The horizontal asymptote is y = 2, the ratio of the leading coefficients.

In general, for y = (ax + b)/(cx + d) with c ≠ 0 and no common factor:

VA: x = −d/c; HA: y = a/c;
x-intercept: x = −b/a (if a ≠ 0); y-intercept: b/d (if d ≠ 0);
domain {x ∈ ℝ | x ≠ −d/c}; range {y ∈ ℝ | y ≠ a/c}.

x	10	100	1000	−10	−100	−1000
(2x + 1)/(x − 3)	3	≈ 2.0722	≈ 2.0070	≈ 1.4615	≈ 1.9320	≈ 1.9930

For large positive x the values are a little above 2; for large negative x they are a little below 2.

The graph of y = (2x + 1)/(x − 3) with its asymptotes x = 3 and y = 2 (dashed). The intercepts are (−0.5, 0) and (0, −1/3). The right branch stays above y = 2 and the left branch stays below it.

Behaviour near the asymptotes

Near the vertical asymptote, use a test value on each side.

x = 3.01: (7.02)/(0.01) = 702, so as x → 3⁺, y → +∞.
x = 2.99: (6.98)/(−0.01) = −698, so as x → 3⁻, y → −∞.

Near the horizontal asymptote, the table shows the right end is above y = 2 and the left end is below.

Can the graph cross its horizontal asymptote? Try to solve (2x + 1)/(x − 3) = 2. Multiply by x − 3: 2x + 1 = 2x − 6, so 1 = −6, which is impossible. So this graph never touches y = 2. That is true for every function of this type (unless it is just a line with a hole). In the lesson on sketching you will meet other rational functions that do cross their horizontal asymptotes.

Worked example

Example 1: All the features of y = (2x + 1)/(x − 3)

List the asymptotes, intercepts, domain and range, and the behaviour near each asymptote.

VA x = 3 and HA y = 2 (ratio of leading coefficients 2/1).
x-intercept −1/2; y-intercept −1/3.
Domain {x ∈ ℝ | x ≠ 3}; range {y ∈ ℝ | y ≠ 2}.
As x → 3⁺, y → +∞; as x → 3⁻, y → −∞.
As x → +∞, y → 2 from above; as x → −∞, y → 2 from below.

See the graph above: one branch is to the upper right of the asymptotes, the other to the lower left.

Worked example

Example 2: A leading coefficient in the denominator

Analyse y = (x − 4)/(2x + 6).

VA: 2x + 6 = 0 gives x = −3.
HA: ratio of leading coefficients, y = 1/2.
x-intercept 4; y-intercept (−4)/6 = −2/3.
Near x = −3: at x = −2.9, (−6.9)/(0.2) = −34.5, so as x → −3⁺, y → −∞. At x = −3.1, (−7.1)/(−0.2) = 35.5, so as x → −3⁻, y → +∞.
Domain {x ∈ ℝ | x ≠ −3}; range {y ∈ ℝ | y ≠ 1/2}.

VA x = −3, HA y = 1/2, intercepts (4, 0) and (0, −2/3). The branches are upper-left and lower-right of the asymptotes.

Worked example

Example 3: A negative leading coefficient

Analyse y = (3 − x)/(x + 1).

Rewrite the numerator in standard order: (−x + 3)/(x + 1).
VA x = −1; HA y = −1/1, so y = −1.
x-intercept 3; y-intercept 3/1 = 3.
Test x = 0 (to the right of the VA): y = 3 > −1, so the right branch is above y = −1. Test x = −2: (5)/(−1) = −5 < −1, so the left branch is below.

VA x = −1, HA y = −1, intercepts (3, 0) and (0, 3).

Left: y = (x − 4)/(2x + 6) with asymptotes x = −3 and y = 0.5. Right: y = (3 − x)/(x + 1) with asymptotes x = −1 and y = −1. Each branch sits in one 'corner' made by the two asymptotes.

Worked example

Example 4: A common factor makes a hole

Describe the graph of y = (x − 2)/(2x − 4).

Factor the denominator: 2x − 4 = 2(x − 2).
For x ≠ 2, y = (x − 2)/(2(x − 2)) = 1/2.
At x = 2 both numerator and denominator are 0, so the function is undefined there, but the graph does not shoot up or down. There is just a missing point.

The graph is the horizontal line y = 1/2 with a hole at (2, 1/2). There is no vertical asymptote. Always look for a common factor before you name the asymptotes.

When the same factor cancels from the top and bottom, the graph has a single missing point (an open circle) instead of an asymptote.

Worked example

Example 5: The batting average model

A(x) = (3 + x)/(10 + x) gives the batting average after x more hits in a row. Find its asymptotes and explain what each means.

Write it as (x + 3)/(x + 10). VA: x = −10. HA: y = 1/1 = 1.
The VA x = −10 has no meaning here, because x counts at-bats and cannot be negative.
The HA A = 1 means the average gets closer and closer to a perfect 1.000, but never reaches it.
A(0) = 0.3 is the current average (the y-intercept).

VA x = −10 (outside the context), HA A = 1 (a perfect average, never quite reached).

Just for fun

If you start a season 0 for 10, you can get a hit in every at-bat for the rest of your life, and your average will approach 1.000 but never reach it. Asymptotes: mathematically proven to keep hope alive.

Hands-on activity

Laundry-basket shooting

You need: a laundry basket or a large bowl, 10 balls of scrunched-up scrap paper, paper and a pencil. Clear a space so nothing breakable is nearby.

Stand about 2 m from the basket and take 10 shots. Count your makes, m.
Your shooting percentage if you make the next x shots in a row is P(x) = (m + x)/(10 + x). Write the formula with your own m.
Make a table for x = 0, 5, 10, 20, 50 and 100. Round to three decimal places.
State the vertical and horizontal asymptotes. Which one has meaning here?
Take more shots and see how many in a row you can make. Is your percentage climbing the way the table predicts?

What to look for: with m = 4, the values are 0.400, 0.600, 0.700, 0.800, 0.900 and 0.945 (rounded). The VA x = −10 has no meaning (you cannot take a negative number of shots). The HA P = 1 means 100%, which you approach but never reach, because you already missed some shots.

Your turn

For y = (x + 2)/(x − 4), state the vertical and horizontal asymptotes.

For y = (x + 2)/(x − 4), find both intercepts.

For y = (x + 2)/(x − 4), state the domain and range.

Describe the behaviour of y = (x + 2)/(x − 4) as x → 4⁺ and as x → 4⁻.

Is the graph of y = (x + 2)/(x − 4) above or below its horizontal asymptote when x is large and positive? Use x = 104.

Find the asymptotes of y = (3x − 6)/(2x + 1).

More practice: warm-up

State the horizontal asymptote of y = (5x − 1)/(x + 2).

State the vertical asymptote of y = (5x − 1)/(x + 2).

Find the x-intercept of y = (x − 6)/(x + 1).

Find the y-intercept of y = (x − 6)/(x + 1).

State the horizontal asymptote of y = (x + 3)/(4x − 1).

More practice: on your own

Find the asymptotes and intercepts of y = (4x − 2)/(2x + 5).

Find the asymptotes and intercepts of y = (−2x + 6)/(x − 1).

Describe the behaviour of y = 3x/(x − 2) near its vertical asymptote.

Does y = (x + 5)/(x − 1) ever cross its horizontal asymptote? Show why.

Describe the graph of y = (x + 3)/(2x + 6).

Challenge

Find a and d so that y = (ax + 4)/(x + d) has vertical asymptote x = −2 and horizontal asymptote y = 3. Then give its x-intercept.

Prove that y = (ax + b)/(cx + d), with c ≠ 0 and ad − bc ≠ 0, never meets its horizontal asymptote y = a/c.

A basketball player has made 12 of 20 free throws. If she makes the next x in a row, her success rate is P(x) = (12 + x)/(20 + x). How many in a row does she need to reach 80%? Can she ever reach 100%?

Check your understanding

Quiz

What is the horizontal asymptote of y = (6x − 1)/(3x + 2)?

What is the x-intercept of y = (2x − 5)/(x + 4)?

What is the y-intercept of y = (2x − 5)/(x + 4)?

Which function has vertical asymptote x = 3 and horizontal asymptote y = −1?

What does the graph of y = (x − 1)/(3x − 3) look like?

As x → 3⁺, what does y = (2x + 1)/(x − 3) do?

For parents and teachers

Big idea (C2.2): for y = (ax + b)/(cx + d) the denominator's zero gives the vertical asymptote, the numerator's zero gives the x-intercept, and the ratio of the leading coefficients a/c gives the horizontal asymptote, because for huge x the constants hardly matter.

How to help: Ask your teen to type (2 × 1000 + 1) ÷ (1000 − 3) into a calculator and explain why the answer is close to 2. Then ask where the graph 'blows up' (x = 3) and why.

Common difficulties:

Using the numerator's zero as the vertical asymptote (the two get swapped).
Writing the HA as y = b/d (the constants) instead of y = a/c (the leading coefficients).
Missing a common factor, which makes a hole instead of an asymptote.
Forgetting that numerators like 3 − x have leading coefficient −1.