Grade by Grade
C. Polynomial and Rational Functions · Rational Functions and Equations

Reciprocals of Linear Functions

C2.1About 70 minutes
Learning goals
  • I can find the vertical and horizontal asymptotes of y = 1/(kx + c).
  • I can state the domain, range, intercepts and positive/negative intervals of the reciprocal of a linear function.
  • I can describe the behaviour of the graph close to a vertical asymptote, using notation such as x → 2⁺.
  • I can sketch y = 1/f(x) from the graph of a linear function f(x), using invariant points.

Words to know

rational function
A function that can be written as f(x) = p(x)/q(x), where p and q are polynomials and q is not the zero polynomial.
reciprocal function
For a function f, the function y = 1/f(x). It is undefined wherever f(x) = 0.
vertical asymptote
A vertical line x = a that the graph approaches as x approaches a: the y-values grow without bound (toward +∞ or −∞).
horizontal asymptote
A horizontal line y = b that the graph approaches as x → +∞ or x → −∞.
invariant point
A point that stays the same under a change. For a reciprocal, the points where f(x) = 1 or f(x) = −1 are invariant, because 1/1 = 1 and 1/(−1) = −1.
x → a⁺ and x → a⁻
Read 'x approaches a from the right' (through values bigger than a) and 'x approaches a from the left' (through values smaller than a).

A trip that never ends

A cyclist rides 60 km. Her travel time is t = 60/v, where v is her speed in km/h. At 30 km/h she needs 2 h. At 6 km/h she needs 10 h. At 0.6 km/h (barely moving) she would need 100 h. The slower she goes, the longer the trip, without any limit. The faster she goes, the closer the time gets to 0, but it never reaches 0.

That 'huge when small, small when huge' pattern is the heart of every rational function, and it is what this unit is about.

Remember from Grade 11: in the MCR3U unit Functions and Equivalent Expressions you met the parent function f(x) = 1/x. Its domain is all real numbers except 0, its range is all real numbers except 0, and it has a vertical asymptote x = 0 and a horizontal asymptote y = 0. You also translated it: y = 1/(x − d) + c has asymptotes x = d and y = c. In this lesson you will go one step further and take the reciprocal of any linear function, then explain every feature of the graph from the line itself.

Reciprocal of a function

If f(x) is a function, its reciprocal is g(x) = 1/f(x). Each y-value of g is just 1 divided by the matching y-value of f.

Take f(x) = x − 2. Then g(x) = 1/(x − 2). Here is how each feature of g comes from the line f.

  • Vertical asymptote. f(2) = 0, and you cannot divide by 0. So g is undefined at x = 2. Near x = 2, f(x) is a tiny number, and 1 divided by a tiny number is huge. The graph shoots up or down along the line x = 2.
  • Horizontal asymptote. When x is far from 2 (very large or very negative), f(x) is huge, and 1 divided by a huge number is close to 0. So the graph hugs the line y = 0 at both ends.
  • No x-intercept. A fraction is 0 only when its numerator is 0. The numerator is 1, so g(x) is never 0.
  • y-intercept. g(0) = 1/(0 − 2) = −1/2.
  • Invariant points. Where f(x) = 1, g(x) = 1/1 = 1 too. Where f(x) = −1, g(x) = −1. For this line, f(3) = 1 and f(1) = −1, so (3, 1) and (1, −1) are on both graphs.
  • Same sign. 1/f(x) has the same sign as f(x). The line is negative for x < 2 and positive for x > 2, so g is too.
f(x) = x − 2 and its reciprocal g(x) = 1/(x − 2)xy−2−1123456−5−4−3−2−11234x = 2f(x) = x − 2g(x) = 1/(x − 2)(3, 1)(1, −1)(0, −0.5)
The dashed line is f(x) = x − 2; the solid curve is its reciprocal. The curves meet at the invariant points (3, 1) and (1, −1). Where the line crosses the x-axis (x = 2), the reciprocal has a vertical asymptote.

Behaviour near the vertical asymptote

Look at the numbers. They show what the graph does just to the left and just to the right of x = 2.

We write x → 2⁺ for 'x approaches 2 from the right' and x → 2⁻ for 'x approaches 2 from the left'.

  • As x → 2⁺, x − 2 is a tiny positive number, so g(x) → +∞ (the graph rises without bound).
  • As x → 2⁻, x − 2 is a tiny negative number, so g(x) → −∞ (the graph falls without bound).
  • As x → +∞, g(x) → 0 from above. As x → −∞, g(x) → 0 from below.
x1.91.991.99922.0012.012.1
x − 2−0.1−0.01−0.00100.0010.010.1
1/(x − 2)−10−100−1000undefined100010010
Tiny values of x − 2 make huge values of the reciprocal. The sign of x − 2 decides whether the graph goes up or down.

Increasing and decreasing

The line f(x) = x − 2 is increasing. As f gets bigger, 1/f gets smaller (on each side of the asymptote, where f keeps the same sign). So g is decreasing on (−∞, 2) and on (2, ∞).

In general, for the reciprocal of a non-constant linear function:

  • if the line is increasing (positive slope), the reciprocal is decreasing on each interval of its domain;
  • if the line is decreasing (negative slope), the reciprocal is increasing on each interval of its domain.

Careful: we say 'decreasing on (−∞, 2) and on (2, ∞)', not 'decreasing on its whole domain'. Jumping across the asymptote, the value goes from very negative to very positive.

Reading the reciprocal from the originalf(x) = 0→ vertical asymptotef(x) = ±1→ invariant point|f(x)| huge→ 1/f(x) near 0f(x) near 0→ 1/f(x) huge
Four ways to read the reciprocal straight from the original function: zeros become asymptotes, ±1 stays put, big becomes small and small becomes big.

Summary: y = 1/(kx + c), with k ≠ 0

  • Vertical asymptote: x = −c/k (the zero of the line).
  • Horizontal asymptote: y = 0.
  • Domain: {x ∈ ℝ | x ≠ −c/k}. Range: {y ∈ ℝ | y ≠ 0}.
  • x-intercept: none. y-intercept: 1/c (if c ≠ 0).
  • Sign: same as the line on each side of the asymptote.
  • Invariant points: where kx + c = 1 or kx + c = −1.
  • Increasing/decreasing: opposite to the line, on each interval.
Worked example

Example 1: All the key features of y = 1/(2x + 6)

For g(x) = 1/(2x + 6), find the asymptotes, domain, range, intercepts, positive and negative intervals, intervals of increase or decrease, and the invariant points.

  1. The line is f(x) = 2x + 6. Its zero: 2x + 6 = 0, so x = −3. Vertical asymptote: x = −3.
  2. The numerator is a constant and the denominator grows without bound, so the horizontal asymptote is y = 0.
  3. Domain {x ∈ ℝ | x ≠ −3}; range {y ∈ ℝ | y ≠ 0}.
  4. No x-intercept. y-intercept: g(0) = 1/6.
  5. The line is positive for x > −3 and negative for x < −3, so g(x) > 0 on (−3, ∞) and g(x) < 0 on (−∞, −3).
  6. The line has slope 2 (increasing), so g is decreasing on (−∞, −3) and on (−3, ∞).
  7. Invariant points: 2x + 6 = 1 gives x = −2.5, so (−2.5, 1). 2x + 6 = −1 gives x = −3.5, so (−3.5, −1).

VA x = −3, HA y = 0, y-intercept 1/6, no x-intercept, positive for x > −3, negative for x < −3, decreasing on both intervals, invariant points (−2.5, 1) and (−3.5, −1).

Worked example

Example 2: A line with a negative slope

Analyse g(x) = 1/(4 − x), including its behaviour near the vertical asymptote.

  1. Zero of the line: 4 − x = 0, so x = 4. VA: x = 4. HA: y = 0.
  2. y-intercept: 1/4. No x-intercept.
  3. 4 − x > 0 when x < 4, so g is positive on (−∞, 4) and negative on (4, ∞).
  4. Near x = 4: as x → 4⁻ (for example x = 3.99), 4 − x = 0.01 is a small positive number, so g(x) → +∞. As x → 4⁺ (x = 4.01), 4 − x = −0.01, so g(x) → −∞.
  5. The line has slope −1 (decreasing), so g is increasing on (−∞, 4) and on (4, ∞).
  6. Invariant points: 4 − x = 1 gives (3, 1); 4 − x = −1 gives (5, −1).

VA x = 4, HA y = 0, y-intercept 1/4; g(x) → +∞ as x → 4⁻ and g(x) → −∞ as x → 4⁺; increasing on each interval.

y = 1/(2x + 6)xy−6−5−4−3−2−112−4−3−2−1123x = −3y = 2x + 6(−2.5, 1)(−3.5, −1)y = 1/(4 − x)xy−11234567−4−3−2−1123x = 4y = 4 − x(3, 1)(5, −1)
Left: the line y = 2x + 6 (dashed) and its reciprocal. Right: the line y = 4 − x (dashed) and its reciprocal. A rising line gives a falling reciprocal, and a falling line gives a rising reciprocal.
Worked example

Example 3: A numerator other than 1

Sketch y = 3/(x + 1) and state its key features.

  1. This is 3 × 1/(x + 1): a vertical stretch by a factor of 3 of the reciprocal of x + 1.
  2. VA: x = −1. HA: y = 0. Domain {x ∈ ℝ | x ≠ −1}, range {y ∈ ℝ | y ≠ 0}.
  3. y-intercept: 3/1 = 3. No x-intercept.
  4. Helpful points: x = 2 gives 3/3 = 1; x = −2 gives 3/(−1) = −3; x = −4 gives 3/(−3) = −1.
  5. Positive for x > −1, negative for x < −1; decreasing on (−∞, −1) and on (−1, ∞).

Two branches in the 'top right' and 'bottom left' of the asymptotes x = −1 and y = 0, passing through (0, 3), (2, 1), (−2, −3) and (−4, −1).

Worked example

Example 4: From features to an equation

The reciprocal of a linear function f(x) = kx + c has a vertical asymptote x = −4 and a y-intercept of 1/8. Find the equation.

  1. The asymptote is the zero of the line, so f(−4) = 0: −4k + c = 0, so c = 4k.
  2. The y-intercept of 1/f is 1/c, so 1/c = 1/8 and c = 8.
  3. Then k = c/4 = 2.

y = 1/(2x + 8). Check: 2(−4) + 8 = 0 ✓ and 1/(0 + 8) = 1/8 ✓.

Just for fun

An asymptote is like the last butter tart on the plate at a family gathering: everyone gets closer and closer to it, but nobody ever actually takes it.

Hands-on activity

Sharing a snack: a reciprocal you can count

You need: 24 crackers (or dried pasta pieces, or coins), paper, a pencil and graph paper.

Two grown-ups always share the snack with n teens. Each person gets c(n) = 24/(n + 2) pieces.

  1. Make a table for n = 0, 1, 2, 4, 6 and 10. For each n, actually deal the 24 pieces into n + 2 equal piles where you can (n = 0, 1, 2, 4, 6, 10 give 2, 3, 4, 6, 8 and 12 piles), and record the pile size.
  2. Plot the points (n, c) on graph paper and join them with a smooth curve.
  3. Use algebra: what would happen at n = −2? Why does that value have no meaning in this situation?
  4. What happens to c(n) as n gets very large? Which asymptote is that?
  5. Is c increasing or decreasing? Explain using the 'big becomes small' idea.

What to look for: c = 12, 8, 6, 4, 3 and 2. The formula has a vertical asymptote at n = −2 (dividing by 0), but n cannot be negative, so in context the graph only shows the right branch. As n grows, c → 0 (the horizontal asymptote c = 0): everyone gets crumbs. c is decreasing because n + 2 is increasing.

Your turn

1

For y = 1/(x + 5), state the equation of the vertical asymptote.

2

For y = 1/(x + 5), state the horizontal asymptote.

3

For y = 1/(x + 5), state the domain and range in set notation.

4

For y = 1/(x + 5), find the intercepts.

5

For y = 1/(x + 5), where is the function positive and where is it negative?

6

For y = 1/(x + 5), describe the behaviour as x → −5⁺ and as x → −5⁻.

7

For y = 1/(x + 5), find the invariant points.

More practice: warm-up

1

State the vertical asymptote of y = 1/(x − 7).

2

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

3

Find the y-intercept of y = 1/(x + 4).

4

Does y = 1/(x − 1) have an x-intercept?

5

If f(5) = 0.2, what is the value of 1/f(x) at x = 5?

More practice: on your own

1

Find the vertical asymptote, y-intercept, positive interval and intervals of increase or decrease for y = 1/(3x − 6).

2

Analyse y = −1/(x + 2): asymptotes, y-intercept, where it is positive, and whether it is increasing or decreasing.

3

Describe the behaviour of y = 1/(5 − x) near its vertical asymptote.

4

Find the invariant points of y = 1/(2x − 3) and its line y = 2x − 3.

5

The graph of y = 1/(kx + c) has vertical asymptote x = 1.5 and y-intercept −1/3. Find k and c.

Challenge

1

The reciprocal of f(x) = kx + c has vertical asymptote x = 3 and passes through (5, −1/4). Find f(x).

2

Explain why the range of y = 1/(kx + c), k ≠ 0, is every real number except 0.

3

A cyclist rides 60 km with a steady tailwind that adds 10 km/h to her speed. If v is her speed in still air, her time is t(v) = 60/(v + 10). (a) What speed v gives t = 2 h? (b) Explain why t is less than 6 h for every possible v. (c) Which asymptote of the formula has no meaning in this context?

Check your understanding

Quiz

1.

What is the vertical asymptote of y = 1/(4x − 2)?

2.

Which statement about y = 1/(x + 3) is true?

3.

A linear function f has f(4) = −1. Which point must be on the graph of y = 1/f(x)?

4.

As x → 1⁺, what happens to y = 1/(x − 1)?

5.

What is the domain of y = 1/(2x + 10)?

6.

The line f(x) = −3x + 1 is decreasing. How does y = 1/(−3x + 1) behave?

For parents and teachers

Big idea (C2.1): the reciprocal of a linear function, y = 1/(kx + c), can be read entirely from the line: the line's zero becomes a vertical asymptote, the reciprocal has the same sign as the line, values of 1 and −1 stay put, and big values become small (and small become big).

How to help: Name a line such as y = x − 3 and ask your teen, 'Where is the vertical asymptote? Where are the invariant points? Is the reciprocal above or below the x-axis when x = 10?' Then check with a calculator: 1 ÷ (10 − 3) ≈ 0.14.

Common difficulties:

  • Thinking the reciprocal has an x-intercept where the line does. That is exactly where the reciprocal has a vertical asymptote.
  • Saying the function is 'decreasing on its domain' when it is decreasing on each of two separate intervals.
  • Mixing up x → 2⁺ (from the right, x slightly bigger than 2) and x → 2⁻ (from the left).
  • Confusing the reciprocal 1/f(x) with the inverse f⁻¹(x) from Grade 11. They are different ideas.