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C. Polynomial and Rational Functions · Rational Functions and Equations

Rational Functions y = (ax + b)/(cx + d): Transformations, Intervals and Equations

C2.2 C2.3About 70 minutes
Learning goals
  • I can rewrite (ax + b)/(cx + d) in the form y = k/(x − h) + q and read its asymptotes and transformations.
  • I can decide whether the function is increasing or decreasing on each interval of its domain.
  • I can find the positive and negative intervals with a sign chart.
  • I can write the equation of a rational function from its asymptotes and a point or intercept.

Words to know

transformed form
y = k/(x − h) + q: the graph of y = 1/x stretched vertically by a factor of |k| (and reflected if k < 0), then translated h units horizontally and q units vertically. Its asymptotes are x = h and y = q.
critical value
An x-value where a function can change sign: a zero of the numerator (x-intercept) or a zero of the denominator (vertical asymptote).
test value
A convenient x-value inside an interval, used to find the sign of the function on that whole interval.
mapping rule
A rule such as (x, y) → (x + h, ky + q) that moves each point of the parent graph to the transformed graph.

The same graph in disguise

In the MCR3U unit Functions and Equivalent Expressions you sketched y = 1/(x − d) + c by moving the asymptotes of y = 1/x. In the last lesson you analysed y = (2x + 1)/(x − 3) by finding its zeros and its leading coefficients. These are the same family. A little division turns one form into the other, and each form makes different features easy to see.

Rewriting by division

Ask: 'How many times does x − 3 go into 2x + 1?' Twice, because 2(x − 3) = 2x − 6. The leftover (remainder) is (2x + 1) − (2x − 6) = 7. So

2x + 1 = 2(x − 3) + 7, and therefore

y = (2x + 1)/(x − 3) = 2 + 7/(x − 3).

Now you can read the graph as a transformation of y = 1/x:

  • a vertical stretch by a factor of 7,
  • a translation 3 units right and 2 units up.

The asymptotes move with the graph: x = 3 and y = 2, just as you found last lesson. The mapping rule is (x, y) → (x + 3, 7y + 2).

y = 7/xxy−4−22468−4−22468(1, 7)(7, 1)(−7, −1)(−1, −7)y = 7/(x − 3) + 2xy24681012−2246810x = 3y = 2(4, 9)(10, 3)(−4, 1)(2, −5)
Left: y = 7/x. Right: the same curve moved 3 right and 2 up, which is y = 7/(x − 3) + 2 = (2x + 1)/(x − 3). Each labelled point moves by (x, y) → (x + 3, y + 2).

Increasing or decreasing?

In y = k/(x − h) + q, the sign of k decides the direction:

  • k > 0: like y = 1/x, the graph is decreasing on (−∞, h) and on (h, ∞);
  • k < 0: the graph is reflected, so it is increasing on each interval.

For y = 2 + 7/(x − 3), k = 7 > 0, so it is decreasing on (−∞, 3) and on (3, ∞).

The range is every real number except the horizontal asymptote: {y ∈ ℝ | y ≠ 2}. A function of this type takes every other value exactly once.

k > 0: y = 2 + 3/(x − 1) fallsxy−4−3−2−1123456−4−2246x = 1y = 2k < 0: y = 2 − 3/(x − 1) risesxy−4−3−2−1123456−4−2246x = 1y = 2
Both graphs have asymptotes x = 1 and y = 2. With k = 3 the branches fall from left to right; with k = −3 they rise.

Positive and negative intervals: the sign chart

A rational function can only change sign at a critical value: a zero of the numerator or a zero of the denominator. For y = (2x + 1)/(x − 3) these are x = −1/2 and x = 3. They cut the number line into three intervals. Pick a test value in each:

  • x = −1: (−1)/(−4) > 0, so positive on (−∞, −1/2);
  • x = 0: (1)/(−3) < 0, so negative on (−1/2, 3);
  • x = 4: (9)/(1) > 0, so positive on (3, ∞).
Sign of (2x + 1)/(x − 3)2x + 1−++x − 3−−+sign+−+−1/2(zero)3(undefined)
The sign chart for (2x + 1)/(x − 3). At x = −1/2 the function equals 0, while at x = 3 it is undefined (dashed). That difference will matter when you solve inequalities.
Worked example

Example 1: Rewriting when c ≠ 1

Write y = (x − 4)/(2x + 6) in the form y = q + k/(x − h). Then state whether it is increasing or decreasing, and its positive and negative intervals.

  1. Divide: (1/2)(2x + 6) = x + 3, and x − 4 = (x + 3) − 7, so x − 4 = (1/2)(2x + 6) − 7.
  2. So y = 1/2 − 7/(2x + 6). Factor 2 from the denominator: y = 1/2 − 3.5/(x + 3).
  3. Asymptotes x = −3 and y = 1/2. Here k = −3.5 < 0, so the function is increasing on (−∞, −3) and on (−3, ∞).
  4. Critical values: −3 (undefined) and 4 (zero). Test x = −4: (−8)/(−2) = 4 > 0. Test x = 0: −4/6 < 0. Test x = 5: 1/16 > 0.
  5. Positive on (−∞, −3) ∪ (4, ∞); negative on (−3, 4).

y = 1/2 − 3.5/(x + 3): increasing on each interval, positive on (−∞, −3) ∪ (4, ∞), negative on (−3, 4).

Worked example

Example 2: A negative leading coefficient

Write y = (3 − x)/(x + 1) in transformed form and describe it.

  1. How many times does x + 1 go into −x + 3? −1 times: −1(x + 1) = −x − 1. Remainder: (−x + 3) − (−x − 1) = 4.
  2. So y = −1 + 4/(x + 1).
  3. Asymptotes x = −1 and y = −1; k = 4 > 0, so it is decreasing on each interval.
  4. Range {y ∈ ℝ | y ≠ −1}.

y = 4/(x + 1) − 1: the graph of y = 1/x stretched by 4, moved 1 left and 1 down.

Worked example

Example 3: Equation from the asymptotes and an intercept

Find an equation of the form y = (ax + b)/(x + d) with vertical asymptote x = 2, horizontal asymptote y = 3 and x-intercept 1.

  1. VA x = 2 means the denominator is x − 2, so d = −2.
  2. HA y = 3 means a/1 = 3, so a = 3: y = (3x + b)/(x − 2).
  3. x-intercept 1: 3(1) + b = 0, so b = −3.

y = (3x − 3)/(x − 2). Check: y-intercept (−3)/(−2) = 3/2, and the graph is consistent with the asymptotes.

Worked example

Example 4: Equation from the asymptotes and a point

A graph has asymptotes x = −1 and y = 2 and passes through (0, 5). Find its equation.

  1. Use the transformed form: y = k/(x + 1) + 2.
  2. Substitute (0, 5): 5 = k/1 + 2, so k = 3.
  3. Combine into one fraction: 3/(x + 1) + 2 = (3 + 2(x + 1))/(x + 1) = (2x + 5)/(x + 1).

y = 3/(x + 1) + 2 = (2x + 5)/(x + 1)

Just for fun

In Grade 3, a remainder was the leftover cookie nobody could share. In Grade 12, the remainder 7 in y = 2 + 7/(x − 3) tells you the stretch factor and whether the graph rises or falls. Leftovers have come a long way.

Hands-on activity

Guess my rational function

You need: paper, a pencil, a calculator and a partner (a family member or friend).

  1. Your partner secretly picks three numbers: a vertical asymptote h, a horizontal asymptote q (not 0) and an x-intercept r (with r ≠ h). They write them down, hidden.
  2. They tell you only the three numbers. You build y = (qx − qr)/(x − h) and simplify.
  3. Check your answer together with a calculator: substitute x = r (you should get 0), and x = h + 0.001 (you should get a number very far from 0, either very large and positive or very large and negative).
  4. Rewrite your function in the form y = k/(x − h) + q. Is it increasing or decreasing?
  5. Swap roles and play three rounds.

What to look for: for h = 2, q = 3, r = 1 the function is y = (3x − 3)/(x − 2) = 3 + 3/(x − 2), decreasing on each interval. In general k = q(h − r): when q and h − r have the same sign, k > 0 and the function is decreasing.

Your turn

1

Write y = (3x + 2)/(x + 1) in the form y = q + k/(x − h).

2

State the asymptotes of y = (3x + 2)/(x + 1) from your transformed form.

3

Is y = (3x + 2)/(x + 1) increasing or decreasing? Explain.

4

Find the positive and negative intervals of y = (3x + 2)/(x + 1).

5

State the range of y = (3x + 2)/(x + 1).

6

Find an equation y = (ax + b)/(x + d) with vertical asymptote x = −4, horizontal asymptote y = −2 and y-intercept 1.

More practice: warm-up

1

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

2

Is y = 4/(x + 2) − 1 increasing or decreasing on each interval?

3

Is y = −3/(x − 5) increasing or decreasing on each interval?

4

State the range of y = 6/(x + 1) − 4.

5

Rewrite (x + 1)/x in the form q + k/x.

More practice: on your own

1

Rewrite y = (4x − 3)/(x − 2) in transformed form and state whether it is increasing or decreasing.

2

Find the positive and negative intervals of y = (2x + 7)/(x + 3).

3

Find an equation y = (ax + b)/(cx + d) with vertical asymptote x = 1/2, horizontal asymptote y = 2 and x-intercept 3.

4

A graph has asymptotes x = 3 and y = −1 and passes through (4, 1). Find its equation as a single fraction.

5

Under y = 4/(x + 2) − 1, where do the points (1, 1) and (−1, −1) of y = 1/x go?

Challenge

1

Show that (ax + b)/(cx + d) = a/c + (bc − ad)/(c(cx + d)), and use it to explain when the function is increasing.

2

For which values of b is y = (2x + b)/(x − 1) increasing on each interval of its domain?

3

A function y = (ax + b)/(x + d) has vertical asymptote x = −2 and passes through (0, 2) and (1, 1.5). Find a, b and d, and state the horizontal asymptote.

Check your understanding

Quiz

1.

Which is equivalent to y = (2x + 1)/(x − 3)?

2.

How does y = −5/(x + 1) + 3 behave?

3.

What is the range of y = (4x − 1)/(2x + 3)?

4.

Which function has vertical asymptote x = 2, horizontal asymptote y = 1 and x-intercept −3?

5.

Where is y = (x − 4)/(x + 1) negative?

6.

Under y = 3/(x − 2) + 5, where does the point (1, 1) of y = 1/x go?

For parents and teachers

Big idea (C2.2, C2.3): every function y = (ax + b)/(cx + d) (without a common factor) is just y = 1/x stretched and moved. Dividing to get y = q + k/(x − h) shows the asymptotes x = h and y = q at a glance, and the sign of k tells you whether the branches rise or fall. A sign chart of the critical values gives the positive and negative intervals.

How to help: Give your teen a pair of asymptotes and a point (for example x = 1, y = 2, through (2, 5)) and ask for the equation (k = 3, so y = 3/(x − 1) + 2). Then ask them to check it by substituting the point.

Common difficulties:

  • Division errors, especially with a negative leading coefficient or with c ≠ 1. Encourage checking by expanding back.
  • Saying 'increasing on its domain' instead of 'increasing on each interval'.
  • Forgetting to include the vertical asymptote as a critical value in the sign chart.