Rational Functions with Quadratics

You already know how to graph a linear expression over a linear expression from graphs of rational functions. This page puts a quadratic into the fraction, either on the bottom or on the top. A quadratic on the bottom can give two vertical asymptotes (or none). A quadratic on the top gives a slanted oblique asymptote instead of a horizontal one. In every case, IB sketches must show all asymptotes and all intercepts with the axes.

Key ideas

Two forms

\text{Type 1: } f(x) = \frac{ax + b}{cx^2 + dx + e} \qquad\qquad \text{Type 2: } f(x) = \frac{ax^2 + bx + c}{dx + e}

The reciprocal function of a linear or quadratic is a special case of these (numerator $1$ ).

Vertical asymptotes

The graph has a vertical asymptote $x = k$ wherever the denominator is $0$ and the numerator isn’t. For Type 1, the denominator is a quadratic, so check its discriminant $d^2 - 4ce$ :

Discriminant of the denominator	Vertical asymptotes
positive	two
zero	one (both sides of it go the same way, because the squared factor doesn’t change sign)
negative	none: the graph is one unbroken curve

If the numerator is also $0$ at a zero of the denominator, the factor cancels and you get a hole instead of an asymptote.

Horizontal asymptote (Type 1)

When the denominator has the higher degree, the fraction gets close to $0$ for large $\lvert x \rvert$ :

\frac{ax + b}{cx^2 + dx + e} \approx \frac{ax}{cx^2} = \frac{a}{cx} \to 0

So the horizontal asymptote is $y = 0$ , the $x$ -axis. Unlike the linear-over-linear case, the graph can cross this asymptote: it does so at its $x$ -intercept, $x = -\dfrac{b}{a}$ .

Oblique asymptote (Type 2)

When the numerator’s degree is one more than the denominator’s, use polynomial division:

\frac{ax^2 + bx + c}{dx + e} = mx + k + \frac{r}{dx + e}

For large $\lvert x \rvert$ the remainder fraction gets close to $0$ , so the graph gets close to the line $y = mx + k$ . That line is the oblique (slant) asymptote. There’s no horizontal asymptote.

The sign of $\dfrac{r}{dx + e}$ tells you which side of the line the graph is on: above the line where the fraction is positive, below where it’s negative.

Intercepts

$y$ -intercept: $f(0)$ , if the denominator isn’t $0$ at $x = 0$ .
$x$ -intercepts: where the numerator is $0$ . For Type 2 the numerator is a quadratic, so there may be two, one or none.

Behaviour near a vertical asymptote

Near $x = k$ , the numerator is close to a fixed non-zero number and the denominator is tiny, so $f(x)$ is huge. Test a value just to the left and just to the right of $k$ to see whether the graph goes up or down on each side. A sign chart of the factors does the same job for every interval at once.

Steps for sketching

Factor the numerator and denominator (cancel any common factor and note the hole).
Find the vertical asymptotes, and the horizontal or oblique asymptote. Draw them as dashed lines.
Find and plot the intercepts.
Work out the sign of $f(x)$ on each interval (or test points near each asymptote).
Draw each branch through its points, approaching the asymptotes. Use your GDC to check the shape and to find any turning points.

Worked examples

Example 1: Two vertical asymptotes

Sketch $f(x) = \dfrac{x + 1}{x^2 - x - 6}$ , showing all asymptotes and intercepts.

Solution. Factor the denominator: $x^2 - x - 6 = (x - 3)(x + 2)$ .

Vertical asymptotes: $x = 3$ and $x = -2$ (the numerator is $4$ and $-1$ there, not $0$ ).
Horizontal asymptote: $y = 0$ , since the denominator has the higher degree.
$x$ -intercept: $x + 1 = 0$ , so $(-1, 0)$ .
$y$ -intercept: $f(0) = \dfrac{1}{-6}$ , so $\left(0, -\dfrac{1}{6}\right)$ .

Sign chart. The sign can only change at $-2$ , $-1$ and $3$ :

Interval	$x + 1$	$x - 3$	$x + 2$	$f(x)$
$x \lt -2$	$-$	$-$	$-$	$-$
$-2 \lt x \lt -1$	$-$	$-$	$+$	$+$
$-1 \lt x \lt 3$	$+$	$-$	$+$	$-$
$x \gt 3$	$+$	$+$	$+$	$+$

So: on the far left the graph is just below the $x$ -axis and drops down beside $x = -2$ . The middle branch comes down from the top beside $x = -2$ , crosses the $x$ -axis at $(-1, 0)$ (crossing the horizontal asymptote!), passes through $\left(0, -\frac{1}{6}\right)$ and drops down beside $x = 3$ . The right branch comes down from the top beside $x = 3$ and levels off just above the $x$ -axis.

y = \dfrac{x + 1}{x^2 - x - 6}

: two vertical asymptotes, and the

x

-axis as horizontal asymptote, which the graph crosses at

(-1, 0)

Example 2: No vertical asymptotes

Sketch $g(x) = \dfrac{2x - 4}{x^2 + 4}$ , and use technology to find its range.

Solution.

Vertical asymptotes: none. $x^2 + 4 \gt 0$ for every $x$ (its discriminant is $0 - 16 \lt 0$ ), so the graph is one unbroken curve.
Horizontal asymptote: $y = 0$ .
Intercepts: $x$ -intercept $(2, 0)$ ; $y$ -intercept $g(0) = \dfrac{-4}{4} = -1$ .
Sign: the denominator is always positive, so $g(x)$ has the sign of $2x - 4$ : negative for $x \lt 2$ , positive for $x \gt 2$ .

The GDC “minimum” and “maximum” tools give a local minimum at $(-0.828, -1.21)$ and a local maximum at $(4.83, 0.207)$ (3 s.f.).

So the graph starts just below the $x$ -axis on the far left, dips to its minimum, rises through $(0, -1)$ and $(2, 0)$ to its maximum, and then falls back toward the $x$ -axis from above.

Range: $\{y \in \mathbb{R} \mid -1.21 \le y \le 0.207\}$ . (In exact form, the turning points are at $x = 2 \pm 2\sqrt{2}$ and the range is $-\frac{1 + \sqrt{2}}{2} \le y \le \frac{\sqrt{2} - 1}{2}$ , which you can find later with calculus.)

Example 3: An oblique asymptote

Sketch $h(x) = \dfrac{x^2 - 2x - 3}{x - 2}$ , showing all asymptotes and intercepts.

Solution. Divide. $x^2 - 2x - 3 = x(x - 2) - 3$ , so

h(x) = x - \frac{3}{x - 2}

Vertical asymptote: $x = 2$ .
Oblique asymptote: $y = x$ .
$x$ -intercepts: $x^2 - 2x - 3 = (x - 3)(x + 1) = 0$ , so $(-1, 0)$ and $(3, 0)$ .
$y$ -intercept: $h(0) = \dfrac{-3}{-2} = \dfrac{3}{2}$ .

Near the asymptotes. For $x$ just above $2$ , $-\dfrac{3}{x - 2}$ is large and negative, so the graph plunges down; just below $2$ , it shoots up. For $x \gt 2$ the fraction $-\dfrac{3}{x - 2}$ is negative, so the graph is below the line $y = x$ ; for $x \lt 2$ it is above the line.

y = x - \dfrac{3}{x - 2}

approaches the oblique asymptote

y = x

at both ends.

There’s no horizontal asymptote: the graph follows the slanted line upward on the right and downward on the left.

Example 4: Oblique asymptote with turning points

Let $k(x) = \dfrac{2x^2 + x - 1}{x + 2}$ . Find all asymptotes and intercepts, and use technology to find the turning points.

Solution. Divide $2x^2 + x - 1$ by $x + 2$ : the quotient is $2x - 3$ and the remainder is $5$ . Check: $(x + 2)(2x - 3) + 5 = 2x^2 + x - 6 + 5 = 2x^2 + x - 1$ ✓.

k(x) = 2x - 3 + \frac{5}{x + 2}

Vertical asymptote: $x = -2$ . Oblique asymptote: $y = 2x - 3$ .
$x$ -intercepts: $2x^2 + x - 1 = (2x - 1)(x + 1) = 0$ , so $\left(\frac{1}{2}, 0\right)$ and $(-1, 0)$ .
$y$ -intercept: $k(0) = -\dfrac{1}{2}$ .
Near $x = -2$ : $k(-1.9) = 43.2$ and $k(-2.1) = -57.2$ , so the right branch comes down from the top and the left branch goes down to the bottom.
Side of the oblique asymptote: $\dfrac{5}{x + 2} \gt 0$ for $x \gt -2$ , so the right branch is above $y = 2x - 3$ ; the left branch is below it.

The GDC gives a local minimum at $(-0.419, -0.675)$ on the right branch and a local maximum at $(-3.58, -13.3)$ on the left branch (3 s.f.).

So the right branch is a “U” shape between the vertical asymptote and the oblique asymptote, with its lowest point at $(-0.419, -0.675)$ , and the left branch is an upside-down “U” with its highest point at $(-3.58, -13.3)$ .

Common mistakes

Giving a horizontal asymptote when there’s an oblique one. If the numerator’s degree is one more than the denominator’s, there’s no horizontal asymptote. Divide to find the oblique asymptote instead.

Using only the quotient’s $x$ term. The oblique asymptote is the whole quotient. For $k(x)$ above it’s $y = 2x - 3$ , not $y = 2x$ .

Thinking the graph can’t cross the horizontal asymptote. For Type 1 functions, the graph crosses $y = 0$ at its $x$ -intercept. Asymptotes describe what happens far away, not near the middle.

Missing that a quadratic denominator has no zeros. Check the discriminant before writing down vertical asymptotes. $x^2 + 4 = 0$ has no real solutions, so $\dfrac{2x - 4}{x^2 + 4}$ has no vertical asymptotes at all.

Forgetting to check for a common factor. If the numerator and denominator share a factor, there’s a hole, not an asymptote, at that $x$ -value.

Leaving off intercepts or asymptote equations. IB mark schemes expect every asymptote labelled with its equation and every axis intercept labelled with its coordinates.

Practice

1. (Warm-up) State the equations of all asymptotes of $y = \dfrac{3x - 1}{x^2 - 9}$ , and find its intercepts.

Solution

$x^2 - 9 = (x - 3)(x + 3)$ , and the numerator isn’t $0$ at $\pm 3$ , so the vertical asymptotes are $x = 3$ and $x = -3$ . Horizontal asymptote $y = 0$ .

$x$ -intercept: $3x - 1 = 0$ , so $\left(\frac{1}{3}, 0\right)$ . $y$ -intercept: $\dfrac{-1}{-9} = \dfrac{1}{9}$ , so $\left(0, \frac{1}{9}\right)$ .

2. (Warm-up) Find the equations of the asymptotes of $y = \dfrac{x^2 + 3x + 5}{x + 1}$ .

Solution

Divide: $x^2 + 3x + 5 = (x + 1)(x + 2) + 3$ , so

y = x + 2 + \frac{3}{x + 1}

Vertical asymptote $x = -1$ ; oblique asymptote $y = x + 2$ .

3. (Warm-up) Explain why $y = \dfrac{x + 5}{x^2 + 2x + 3}$ has no vertical asymptotes. What is its horizontal asymptote?

Solution

The denominator has discriminant $2^2 - 4(1)(3) = -8 \lt 0$ , so it’s never $0$ and the function is defined for every real $x$ .

The denominator has the higher degree, so the horizontal asymptote is $y = 0$ .

4. (Core) Sketch $y = \dfrac{2x}{x^2 - 1}$ , showing all asymptotes and intercepts. Describe its symmetry.

Solution

$x^2 - 1 = (x - 1)(x + 1)$ : vertical asymptotes $x = 1$ and $x = -1$ . Horizontal asymptote $y = 0$ . The only intercept is the origin $(0, 0)$ .

Signs (from the factors $2x$ , $x - 1$ , $x + 1$ ): negative for $x \lt -1$ , positive for $-1 \lt x \lt 0$ , negative for $0 \lt x \lt 1$ , positive for $x \gt 1$ .

So the left branch is just below the $x$ -axis and drops beside $x = -1$ ; the middle branch comes down from the top beside $x = -1$ , passes through the origin, and drops beside $x = 1$ ; the right branch comes down from the top beside $x = 1$ and levels off just above the $x$ -axis.

Replacing $x$ by $-x$ gives $\dfrac{-2x}{x^2 - 1} = -y$ , so the function is odd: the graph has rotational symmetry of order $2$ about the origin.

5. (Core) Sketch $y = \dfrac{x^2 - 4}{x + 1}$ , showing all asymptotes and intercepts, and state on which side of the oblique asymptote each branch lies.

Solution

Divide: $x^2 - 4 = (x + 1)(x - 1) - 3$ , so

y = x - 1 - \frac{3}{x + 1}

Vertical asymptote $x = -1$ ; oblique asymptote $y = x - 1$ .

$x$ -intercepts: $x^2 - 4 = 0$ , so $(-2, 0)$ and $(2, 0)$ . $y$ -intercept: $\dfrac{-4}{1} = -4$ , so $(0, -4)$ .

For $x \gt -1$ , $-\dfrac{3}{x + 1} \lt 0$ , so the right branch is below $y = x - 1$ : it comes up from the bottom beside $x = -1$ , passes through $(0, -4)$ and $(2, 0)$ , and approaches the line from below. For $x \lt -1$ , the fraction is positive, so the left branch is above the line: it follows the line on the far left, passes through $(-2, 0)$ , and shoots up beside $x = -1$ .

6. (Core) Find the equations of the asymptotes of $y = \dfrac{3x^2 - x + 2}{2x - 1}$ , and show that the graph has no $x$ -intercepts.

Solution

Divide $3x^2 - x + 2$ by $2x - 1$ . First term: $\dfrac{3}{2}x$ , and $\dfrac{3}{2}x(2x - 1) = 3x^2 - \dfrac{3}{2}x$ , leaving $\dfrac{1}{2}x + 2$ . Next term: $\dfrac{1}{4}$ , and $\dfrac{1}{4}(2x - 1) = \dfrac{1}{2}x - \dfrac{1}{4}$ , leaving $\dfrac{9}{4}$ .

y = \frac{3}{2}x + \frac{1}{4} + \frac{9/4}{2x - 1}

Vertical asymptote $x = \dfrac{1}{2}$ ; oblique asymptote $y = \dfrac{3}{2}x + \dfrac{1}{4}$ .

$x$ -intercepts need $3x^2 - x + 2 = 0$ , whose discriminant is $1 - 24 = -23 \lt 0$ . So there are none.

7. (Core) The graph of $f(x) = \dfrac{ax^2 + bx + 3}{x - 1}$ has oblique asymptote $y = 2x + 5$ . Find $a$ and $b$ , and write $f(x)$ as a linear function plus a fraction.

Solution

The quotient when the numerator is divided by $x - 1$ must be $2x + 5$ , so

ax^2 + bx + 3 = (x - 1)(2x + 5) + r = 2x^2 + 3x - 5 + r

Comparing coefficients: $a = 2$ , $b = 3$ , and $-5 + r = 3$ , so $r = 8$ .

f(x) = \frac{2x^2 + 3x + 3}{x - 1} = 2x + 5 + \frac{8}{x - 1}

8. (Challenge) Let $f(x) = \dfrac{x + k}{x^2 - 4}$ .

(a) For which values of $k$ does the graph have only one vertical asymptote?
(b) Describe the graph when $k = 2$ .

Solution

(a) The denominator is $0$ at $x = \pm 2$ . A zero of the denominator gives a hole instead of an asymptote if the numerator is also $0$ there: $2 + k = 0$ or $-2 + k = 0$ . So $k = -2$ or $k = 2$ .

(b) With $k = 2$ :

f(x) = \frac{x + 2}{(x - 2)(x + 2)} = \frac{1}{x - 2}, \qquad x \ne -2

The graph is the reciprocal graph $y = \dfrac{1}{x - 2}$ (vertical asymptote $x = 2$ , horizontal asymptote $y = 0$ , $y$ -intercept $-\frac{1}{2}$ ) with a hole at $\left(-2, -\frac{1}{4}\right)$ .

9. (Challenge) Let $f(x) = \dfrac{x^2 + 4}{x}$ .

(a) State the equations of the asymptotes.
(b) By writing $y = \dfrac{x^2 + 4}{x}$ as a quadratic equation in $x$ , show that the graph never takes values between $-4$ and $4$ . Hence state the range and the turning points.

Solution

(a) $f(x) = x + \dfrac{4}{x}$ , so the vertical asymptote is $x = 0$ and the oblique asymptote is $y = x$ .

(b) Multiply by $x$ : $yx = x^2 + 4$ , so $x^2 - yx + 4 = 0$ . For a given $y$ , there is a point on the graph at that height only if this quadratic in $x$ has a real solution, which needs

(-y)^2 - 4(1)(4) \ge 0 \quad\Rightarrow\quad y^2 \ge 16 \quad\Rightarrow\quad y \le -4 \text{ or } y \ge 4

So the range is $\{y \in \mathbb{R} \mid y \le -4 \text{ or } y \ge 4\}$ . The extreme heights $y = \pm 4$ happen when the discriminant is $0$ , giving the single solution $x = \dfrac{y}{2}$ : a local minimum at $(2, 4)$ and a local maximum at $(-2, -4)$ . Check: $f(2) = \dfrac{8}{2} = 4$ ✓ and $f(-2) = \dfrac{8}{-2} = -4$ ✓.