Year 11 Extended · Unit 4 Rational expressions and functions · Lesson 5 of 7

Where do the asymptotes go when the reciprocal graph is moved?

The graph of y = k/x can be moved like any other graph. This lesson asks what happens to its asymptotes when it is, and how to read them from y = k/(x − p) + q and from (ax + b)/(cx + d).

FunctionsGDC optional

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 10: the graph of y = x2 is moved 3 right. What is its new rule?
    Answery = (x − 3)2.
  2. From lesson 4: what are the asymptotes of y = 6/x?
    Answerx = 0 and y = 0.
  3. From lesson 3: write 3 + 2/(x − 1) as one fraction.
    Answer(3x − 1) ÷ (x − 1).
Words used in this lesson
translation
moving a graph without turning or stretching it. For example: 3 left and 2 up.
intercept
a point where a graph crosses an axis. For example: where x = 0.
vertical asymptote
a vertical line the graph approaches. For example: x = 0 for y = 1/x.
horizontal asymptote
a horizontal line the graph approaches. For example: y = 0 for y = 1/x.
ratio of the x terms
the number in front of x on top divided by the number in front of x below. For example: 6x over 2x is 3.

What does the graph do when you shift it?

A school in Jaipur plans a trip. The coach costs 6000 rupees, shared equally by the students, and entry, meals and a guide come to 3000 rupees for each student. One teacher travels free, so with x people on the coach the cost per student, in thousands of rupees, is C = 6/(x − 1) + 3. It is the reciprocal graph, moved.

Predict first: the asymptotes of y = 6/x are the two axes. Where will they be for this graph? Use the sliders to move the graph and watch the dashed lines. Which slider moves which line?

2
1
−1
x = 1vertical asymptote
y = −1horizontal asymptote

y = 2/(x − 1) − 1 = (−x + 3)/(x − 1). Vertical asymptote x = 1, horizontal asymptote y = −1.

Show me the steps
  1. k = 2, p = 1, q = −1: y = 2/(x − 1) − 1 = (−x + 3)/(x − 1). The asymptotes are x = 1 and y = −1.
  2. Move p to 3: y = 2/(x − 3) − 1 = (−x + 5)/(x − 3). The graph slides 2 to the right and the vertical asymptote moves to x = 3.
  3. Move q to 2: y = 2/(x − 3) + 2 = (2x − 4)/(x − 3). The graph slides 3 up and the horizontal asymptote moves to y = 2.
  4. Make k negative: y = −2/(x − 3) + 2 = (2x − 8)/(x − 3). The asymptotes stay where they are and the branches swap sides.

y = k/(x − p) + q is y = k/x moved p across and q up. Its asymptotes move with it: the vertical one is x = p and the horizontal one is y = q. The domain is x ≠ p and the range is y ≠ q.

Written as one fraction, (ax + b)/(cx + d) has (when c is not 0 and ad is not bc) the vertical asymptote x = −d/c, where the bottom is zero, and the horizontal asymptote y = a/c, the ratio of the x terms, because the numbers b and d matter less as x gets large.

Mind the sign of p. y = 6/(x + 2) is moved 2 to the left, so its vertical asymptote is x = −2, not x = 2. Test it: the bottom x + 2 is zero when x = −2.

Worked examples

1. For y = 9/(x + 2) − 4 find the asymptotes and both intercepts.

  1. The bottom is zero at x = −2, and the graph is moved 4 down: x = −2 and y = −4.
  2. At x = 0, y = 4.5 − 4 = 0.5. At y = 0, 9/(x + 2) = 4, so x + 2 = 2.25 and x = 0.25.

2. Find the asymptotes of y = (2x + 1)/(x − 3), and write it as k/(x − p) + q.

  1. The bottom is zero at x = 3. For large x the 1 and the −3 hardly matter, so y gets close to 2x/x = 2.
  2. Write 2x + 1 = 2(x − 3) + 7, so y = 2 + 7/(x − 3), with x = 3 and y = 2.

3. Find the domain and range of y = (4x − 5)/(2x + 6).

  1. The bottom is zero at x = −3, and the ratio of the x terms is 4/2 = 2.
  2. Domain: x ≠ −3. Range: y ≠ 2.

Your go. The first step is given. Write y = (3x + 2)/(x + 1) in the form q + k/(x − p) and state its asymptotes. The top is 3(x + 1) − 1.

  1. Split the fraction into 3(x + 1)/(x + 1) and the remainder.
  2. Read p and q from the result.
Check the answer

y = 3 − 1/(x + 1), so the asymptotes are x = −1 and y = 3.

With a GDC (optional)

A GDC can graph a rule such as y = (2x + 1)/(x − 3). Enter it with brackets round the top and the bottom, and use the table to see y get close to 2 for large x, and the trace near x = 3 to see y grow. The graph may show a near-vertical line at the vertical asymptote: that line is only the screen joining two branches, and it is not part of the graph.

Use the intersect or zero tool to read an intercept, then confirm it by hand. Everything on this page can be done without a GDC.

Check before you go on

1. Which graph has the asymptotes x = −4 and y = 3?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. y = 5/(x − 2) + 3. Find y when x = 3.

3. Find the x intercept of y = (6x − 15)/(2x + 1). Give your answer as a decimal.

Level 2 · Building

4. The graph of y = (4x − 3)/(2x + 10) has a vertical asymptote x = c. Find c.

5. The graph of y = a/(x − 1) + 4 passes through the point (3, 7). Find a.

Level 3 · Stretch

6. At what value of x does the graph of y = (3x + 2)/(x − 1) reach the height y = 5? Give your answer as a decimal.

7. The graph of y = (ax + 6) ÷ (2x − 3), where a is positive, never takes the value y = 4. Find its x intercept. Give your answer as a decimal.

Something to think about

Can the graph of y = (ax + b)/(cx + d) ever be a straight line? What would have to be true of a, b, c and d?

One way to think about it

Yes, in two cases: when the top is a multiple of the bottom, for example (2x + 6)/(x + 3) = 2, which is the line y = 2 with a hole at x = −3, or when c = 0, which gives a straight line. Otherwise there is always a vertical asymptote.

Exit ticket

8. The graph of y = 2/x is moved 3 to the left and 5 down. Which is its rule as one fraction?