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C. Polynomial and Rational Functions · Rational Functions and Equations

Reciprocals of Quadratic Functions with Two Zeros

C2.1About 70 minutes
Learning goals
  • I can find the two vertical asymptotes of y = 1/(ax² + bx + c) when the quadratic has two zeros.
  • I can use the vertex of the parabola to find the turning point of the middle branch of the reciprocal.
  • I can state the domain, range, intercepts, positive/negative intervals and increasing/decreasing intervals.
  • I can describe the behaviour near each vertical asymptote and sketch the graph.

Words to know

turning point
A point where a graph changes from increasing to decreasing (a local maximum) or from decreasing to increasing (a local minimum).
local maximum
A point that is higher than all nearby points on the graph. It need not be the highest point of the whole graph.
branch
One connected piece of a graph. The reciprocal of a quadratic with two zeros has three branches.
sign chart
A number line divided at the zeros and asymptotes of a function, showing whether the function is positive or negative in each interval.
even function
A function with f(−x) = f(x) for every x in its domain. Its graph is symmetric in the y-axis.

Three pieces instead of two

The reciprocal of a line has one vertical asymptote and two branches. A parabola can cross the x-axis twice, so its reciprocal can have two vertical asymptotes and three branches. The middle branch is the interesting one: it has a turning point, and that turning point comes straight from the vertex of the parabola.

Remember from Grade 11: in the MCR3U unit Quadratic Functions you found zeros by factoring, found the vertex by completing the square or from the zeros (the axis of symmetry is halfway between them), and used the discriminant b² − 4ac to count zeros. You will use all of those tools here.

The rules from the last lesson still work

Everything you learned for lines still applies, because the reciprocal only cares about the values of f(x):

  • Zeros of f → vertical asymptotes of 1/f.
  • f(x) = ±1 → invariant points.
  • Same sign as f.
  • Large |f| → reciprocal near 0, so the horizontal asymptote is y = 0.
  • f increasing → 1/f decreasing, and the reverse, on each interval where f keeps one sign.

Take f(x) = x² − 4 = (x − 2)(x + 2). Its zeros are ±2, so g(x) = 1/(x² − 4) has vertical asymptotes x = −2 and x = 2.

The new feature is the vertex. The parabola has its minimum at (0, −4). On the middle interval (−2, 2), f is negative, and its most negative value is −4. The reciprocal of −4 is −1/4, and every other value of f there (between −4 and 0) gives a reciprocal below −1/4. So the middle branch has a local maximum at (0, −1/4).

f(x) = x² − 4 and its reciprocal g(x) = 1/(x² − 4)xy−4−3−2−11234−5−4−3−2−11234x = −2x = 2f(x) = x² − 4gvertex (0, −4)(0, −0.25)
The dashed parabola f(x) = x² − 4 and its reciprocal (solid). Zeros at x = ±2 become vertical asymptotes. The vertex (0, −4) becomes the local maximum (0, −0.25). The four dots without labels are the invariant points, where f(x) = 1 or −1.

A sign chart

The sign of 1/f(x) only changes at the zeros of f (here, the asymptotes). Test one value in each interval, or use the signs of the factors. Then you also know how the graph behaves near each asymptote.

  • As x → 2⁺: x − 2 is tiny and positive, x + 2 ≈ 4, so the product is tiny and positive and g(x) → +∞.
  • As x → 2⁻: the product is tiny and negative, so g(x) → −∞.
  • As x → −2⁻: x + 2 is tiny and negative, x − 2 ≈ −4, so the product is tiny and positive: g(x) → +∞.
  • As x → −2⁺: the product is tiny and negative: g(x) → −∞.
Sign of 1/((x + 2)(x − 2))x + 2−++x − 2−−+sign+−+−2(undefined)2(undefined)
Factor signs multiply to give the sign of the reciprocal in each interval: positive outside the asymptotes, negative between them. Both x-values are dashed because the function is undefined there.

Summary: y = 1/(ax² + bx + c) when the quadratic has two zeros r₁ < r₂

  • Vertical asymptotes: x = r₁ and x = r₂. Horizontal asymptote: y = 0.
  • Domain: {x ∈ ℝ | x ≠ r₁, r₂}.
  • Turning point: if the parabola's vertex is (h, k), the middle branch has a turning point at (h, 1/k). When a > 0 it is a local maximum; when a < 0 it is a local minimum.
  • Range: for a > 0, (−∞, 1/k] ∪ (0, ∞); for a < 0, (−∞, 0) ∪ [1/k, ∞).
  • Symmetry: the graph is symmetric about the line x = h, like the parabola.
  • No x-intercepts; y-intercept 1/c (if c ≠ 0).

In this course we write intervals of increase and decrease as open intervals, for example (−2, 0). (Some textbooks include the turning point; both are accepted.)

Worked example

Example 1: All the key features of y = 1/(x² − 4)

State the asymptotes, domain, range, intercepts, positive and negative intervals, and intervals of increase and decrease.

  1. VAs: x = −2 and x = 2. HA: y = 0.
  2. Domain {x ∈ ℝ | x ≠ ±2}.
  3. y-intercept g(0) = 1/(−4) = −1/4. No x-intercepts.
  4. Local maximum (0, −1/4), so the range is (−∞, −1/4] ∪ (0, ∞).
  5. Positive on (−∞, −2) ∪ (2, ∞); negative on (−2, 2).
  6. The parabola decreases for x < 0 and increases for x > 0, so g increases on (−∞, −2) and (−2, 0) and decreases on (0, 2) and (2, ∞).
  7. Since (−x)² − 4 = x² − 4, g(−x) = g(x): the function is even, symmetric in the y-axis.

VAs x = ±2, HA y = 0, y-intercept −1/4, local max (0, −1/4), range (−∞, −1/4] ∪ (0, ∞).

Worked example

Example 2: Finding the turning point from the vertex

Analyse y = 1/(x² − 6x + 5).

  1. Factor: x² − 6x + 5 = (x − 1)(x − 5). VAs: x = 1 and x = 5. HA: y = 0.
  2. Axis of symmetry: halfway between the zeros, x = 3. The parabola's value there is 9 − 18 + 5 = −4, so its vertex is (3, −4).
  3. Turning point of the reciprocal: (3, 1/(−4)) = (3, −1/4), a local maximum because a = 1 > 0.
  4. y-intercept: 1/5.
  5. Positive on (−∞, 1) ∪ (5, ∞); negative on (1, 5).
  6. Increasing on (−∞, 1) and (1, 3); decreasing on (3, 5) and (5, ∞).
  7. Range: (−∞, −1/4] ∪ (0, ∞).

VAs x = 1 and x = 5, local maximum (3, −1/4), y-intercept 1/5.

Worked example

Example 3: A parabola that opens down

Analyse y = 1/(−x² + 2x + 3).

  1. −x² + 2x + 3 = −(x² − 2x − 3) = −(x − 3)(x + 1). Zeros 3 and −1, so VAs x = −1 and x = 3.
  2. Vertex: x = 1 (halfway), and −1 + 2 + 3 = 4, so the vertex is (1, 4), a maximum of the parabola.
  3. Between the zeros the parabola is positive, with largest value 4. So the reciprocal is positive there with smallest value 1/4: a local minimum at (1, 1/4).
  4. y-intercept 1/3. Positive on (−1, 3); negative on (−∞, −1) ∪ (3, ∞).
  5. The parabola increases for x < 1 and decreases for x > 1, so the reciprocal decreases on (−∞, −1) and (−1, 1) and increases on (1, 3) and (3, ∞).
  6. Range: (−∞, 0) ∪ [1/4, ∞).

VAs x = −1 and x = 3, local minimum (1, 1/4), range (−∞, 0) ∪ [1/4, ∞).

y = 1/((x − 1)(x − 5))xy1234561−1x = 1x = 5(3, −0.25)(0, 0.2)y = 1/(−x² + 2x + 3)xy−2−112341−1x = −1x = 3(1, 0.25)
Left: y = 1/((x − 1)(x − 5)) has a middle branch below the x-axis with a local maximum at (3, −0.25). Right: y = 1/(−x² + 2x + 3) has a middle branch above the x-axis with a local minimum at (1, 0.25).
Worked example

Example 4: Writing the equation from a graph

A graph has the form y = a/((x − r)(x − s)). It has vertical asymptotes x = −1 and x = 4, and passes through (0, 1/2). Find its equation.

  1. The asymptotes give the factors: y = a/((x + 1)(x − 4)).
  2. Substitute (0, 1/2): 1/2 = a/((1)(−4)) = −a/4.
  3. So a = −2.

y = −2/((x + 1)(x − 4)), which is −2/(x² − 3x − 4).

Just for fun

The middle branch of y = 1/(x² − 4) peaks at y = −1/4. It is a mountain whose summit is below sea level, and it is not even under water.

Hands-on activity

Build a reciprocal from a parabola, point by point

You need: graph paper, a sharp pencil, a ruler and a calculator.

  1. Draw axes from −5 to 5 on both axes (2 squares per unit works well).
  2. Make a table for f(x) = x² − 4 with x = −4, −3, −2.5, −2, −1.5, −1, 0, 1, 1.5, 2, 2.5, 3, 4. Plot the parabola lightly.
  3. Add a row for 1/f(x), rounded to two decimal places. Write 'undefined' at x = ±2.
  4. Draw dashed lines at x = −2 and x = 2. Plot the reciprocal points and join them into three branches.
  5. Circle the four places where the parabola is at height 1 or −1. Check that the reciprocal passes through them too.

What to look for: f values 12, 5, 2.25, 0, −1.75, −3, −4, ... and reciprocals 0.08, 0.2, 0.44, undefined, −0.57, −0.33, −0.25, ... (the table is symmetric). The middle branch tops out at (0, −0.25). The invariant points are at x = ±√5 ≈ ±2.24 (height 1) and x = ±√3 ≈ ±1.73 (height −1).

Your turn

1

Find the vertical asymptotes of y = 1/(x² − 2x − 8).

2

State the horizontal asymptote of y = 1/(x² − 2x − 8).

3

Find the y-intercept of y = 1/(x² − 2x − 8).

4

Find the turning point of y = 1/(x² − 2x − 8). Is it a maximum or a minimum?

5

Where is y = 1/(x² − 2x − 8) positive? Where is it negative?

6

State the range of y = 1/(x² − 2x − 8).

7

State the intervals of increase and decrease of y = 1/(x² − 2x − 8).

8

Describe the behaviour of y = 1/(x² − 2x − 8) as x → 4⁺ and as x → 4⁻.

More practice: warm-up

1

State the vertical asymptotes of y = 1/((x − 3)(x + 7)).

2

State the horizontal asymptote of y = 1/(x² − 16).

3

Find the y-intercept of y = 1/(x² − 16).

4

The parabola y = x² − 16 has vertex (0, −16). What is the turning point of y = 1/(x² − 16)?

5

State the domain of y = 1/(x² − 1).

More practice: on your own

1

Analyse y = 1/(x² + 2x − 15): asymptotes, y-intercept and turning point.

2

State the asymptotes, y-intercept and range of y = 1/(2x² − 8).

3

Analyse y = 1/(9 − x²): asymptotes, sign, turning point and range.

4

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

5

A graph of the form y = a/((x − r)(x − s)) has vertical asymptotes x = 0 and x = 6 and passes through (3, 1/3). Find its equation.

Challenge

1

For which values of k does y = 1/(x² − 4x + k) have exactly two vertical asymptotes?

2

Prove that y = 1/(x² − 4) is an even function, and explain what that tells you about its graph.

3

Find the range of y = 1/(x² + 4x − 5) and explain why no y-value in (−1/9, 0] is reached.

Check your understanding

Quiz

1.

How many vertical asymptotes does y = 1/(x² − 25) have?

2.

The parabola y = x² − 6x + 8 has vertex (3, −1). What is the turning point of y = 1/(x² − 6x + 8)?

3.

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

4.

What is the range of y = 1/(x² − 4)?

5.

Which describes the turning point of y = 1/(1 − x²)?

6.

As x → 3⁺, what does y = 1/(x² − 9) do?

For parents and teachers

Big idea (C2.1): when a quadratic has two zeros, its reciprocal has two vertical asymptotes and three branches. The vertex (h, k) of the parabola becomes the turning point (h, 1/k) of the middle branch.

How to help: Ask your teen to explain, in words, why the middle branch of 1/(x² − 4) is below the x-axis (the parabola is negative there) and why its highest point is at −1/4 (the parabola's lowest value there is −4). If they can explain that, they understand the lesson.

Common difficulties:

  • Calling the turning point (h, k) instead of (h, 1/k).
  • Getting the range wrong by forgetting that the outer branches cover all positive numbers while the middle branch only reaches down from −1/4.
  • Sign errors near the asymptotes. Encourage a quick sign chart of the factors.
  • Forgetting that a parabola opening down (a < 0) flips everything: the middle branch is above the x-axis and has a local minimum.