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Wize High School Grade 12 Calculus Textbook > Curve Sketching

Asymptotes

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Vertical Asymptotes

The graph of f(x)f\left(x\right) has a vertical asymptote at x=ax=a if the left-hand and/or right-hand limit as x→ax\to a is ∞\infty or −∞-\infty

Wize Tip
Vertical asymptotes are usually denoted by a dotted vertical line x=ax=a on the graph.


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How do we find vertical asymptotes?

Determine any point(s) x=ax=a where the graph is undefined.

Wize Tip
*If we are given a rational function p(x)q(x)\displaystyle \frac{p\left(x\right)}{q\left(x\right)}, we want to check when the denominator equals 0.

Set up the following table to more easily determine the left and right-hand limits
Value of xf(x)f(x)→?x→a−f(x)=...−∞ or +∞x→a+f(x)=...−∞ or +∞⋮⋮⋮\begin{array}{|c|c|c|} \hline \text{Value of }x&f(x)&f(x)\to?\\ \hline x\to a^-&f(x)=...&-\infty\ or\ +\infty\\ \hline x\to a^+&f(x)=...&-\infty\ or\ +\infty\\ \vdots &\vdots&\vdots\\ \hline \end{array}
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Example: Vertical Asymptotes

Find the vertical asymptote(s), if any, of the following functions.
a) f(x)=2xx2−4\displaystyle f(x)=\frac{2x}{x^2-4}

This is a rational function, which is undefined (discontinuous) when the denominator equals 0:
x2−4=0x^2-4=0
(x−2)(x+2)=0\left(x-2\right)\left(x+2\right)=0
x=2 or −2x=2\ or\ -2

Value of xf(x)=2x(x−2)(x+2)f(x)→?x→−2−−4(−4)(0−)−∞x→−2+−4(−4)(0+)+∞x→2−4(0−)(4)−∞x→2+4(0+)(4)+∞\begin{array}{|c|c|c|} \hline \text{Value of }x&f(x)=\frac{2x}{(x-2)(x+2)}&f(x)\to?\\ \hline x\to-2^-&\frac{-4}{(-4)(0^-)}&-\infty\\ \hline x\to-2^+&\frac{-4}{(-4)(0^+)}&+\infty\\ \hline x\to2^-&\frac{4}{(0^-)(4)}&-\infty\\ \hline x\to2^+&\frac{4}{(0^+)(4)}&+\infty\\ \hline \end{array}

Therefore, this graph has vertical asymptotes at x=−2x=-2 and x=2x=2.

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b) g(x)=x+2x2+5x+6\displaystyle g\left(x\right)=\frac{x+2}{x^2+5x+6}

This is a rational function, which is undefined (discontinuous) when the denominator equals 0:
x2+5x+6=0x^2+5x+6=0
(x+2)(x+3)=0\left(x+2\right)\left(x+3\right)=0
x=−2 or −3x=-2\ or\ -3

Value of xf(x)=x+2(x+2)(x+3)f(x)→?x→−3−−1(−1)(0−)+∞x→−3+−1(−1)(0+)−∞x→−2−1x+3=111x→−2+1x+3=111\begin{array}{|c|c|c|} \hline \text{Value of }x&f(x)=\frac{x+2}{(x+2)(x+3)}&f(x)\to?\\ \hline x\to-3^-&\frac{-1}{(-1)(0^-)}&+\infty\\ \hline x\to-3^+&\frac{-1}{(-1)(0^+)}&-\infty\\ \hline x\to-2^-&\frac{1}{x+3}=\frac{1}{1}&1\\ \hline x\to-2^+&\frac{1}{x+3}=\frac{1}{1}&1\\ \hline \end{array}

Therefore, this graph has vertical asymptotes at x=−3x=-3 only (it has a point discontinuity at x=−2x=-2).

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c) h(x)=1x−1+1x2+3x\displaystyle h\left(x\right)=\frac{1}{x-1}+\frac{1}{x^2+3x}

Rewrite:
h(x)=(x2+3x)+(x−1)(x−1)(x2+3x)=x2+4x−1(x−1)(x)(x+3)\displaystyle h\left(x\right)=\frac{\left(x^2+3x\right)+\left(x-1\right)}{\left(x-1\right)\left(x^2+3x\right)}=\frac{x^2+4x-1}{\left(x-1\right)\left(x\right)\left(x+3\right)}

This is a rational functions, which is undefined (discontinuous) when the denominator equals 0:
(x−1)(x)(x+3)=0\left(x-1\right)\left(x\right)\left(x+3\right)=0
x=1, 0,  or−3x=1,\ 0,\ \ or-3

Value of xf(x)=x2+4x−1(x−1)(x)(x+3)f(x)→?x→−3−−4(−4)(−3)(0−)+∞x→−3+−4(−4)(−3)(0+)−∞x→0−−1(−1)(0−)(3)−∞x→0+−1(−1)(0+)(3)+∞x→1−4(0−)(1)(4)−∞x→1+4(0+)(1)(4)+∞\begin{array}{|c|c|c|} \hline \text{Value of }x&f(x)=\frac{x^2+4x-1}{(x-1)(x)(x+3)}&f(x)\to?\\ \hline x\to-3^-&\frac{-4}{(-4)(-3)(0^-)}&+\infty\\ \hline x\to-3^+&\frac{-4}{(-4)(-3)(0^+)}&-\infty\\ \hline x\to0^-&\frac{-1}{(-1)(0^-)(3)}&-\infty\\ \hline x\to0^+&\frac{-1}{(-1)(0^+)(3)}&+\infty\\ \hline x\to1^-&\frac{4}{(0^-)(1)(4)}&-\infty\\ \hline x\to1^+&\frac{4}{(0^+)(1)(4)}&+\infty\\ \hline \end{array}

Therefore, this graph has vertical asymptotes at x=−3x=-3, x=0x=0, and x=1x=1.

Practice: Vertical Asymptotes

Find the vertical asymptote(s), if any, of f(t)=t2−1(t−2)(t2−6t+5)\displaystyle f(t)=\frac{t^2 -1}{(t-2)(t^2-6t+5)}.

[Select all that apply]
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Horizontal Asymptotes

The graph of f(x)f(x) has a horizontal asymptote at y=ky=k if lim⁡x→∞f(x)=k\displaystyle \lim_{x\to\infty}f\left(x\right)=k and/or lim⁡x→−∞f(x)=k\displaystyle \lim_{x\to-\infty}f\left(x\right)=k.

Wize Tip
Horizontal asymptotes are usually denoted by a dotted horizontal line y=ky=k on the graph.


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How do we find horizontal asymptotes?

Determine lim⁡x→∞f(x)\displaystyle \lim_{x\to\infty}f\left(x\right) and lim⁡x→−∞f(x)\displaystyle \lim_{x\to-\infty}f\left(x\right).

For rational functions p(x)q(x)\orange{\bold{\displaystyle \frac{p\left(x\right)}{q\left(x\right)}}}:
  • We divide all terms by the highest degree term and then evaluate the limit
  • We also want to determine if the graph is approaching the horizontal asymptote from above or below
Write it Down
Short-cut for rational functions p(x)q(x)\bold{\displaystyle \frac{p\left(x\right)}{q\left(x\right)}}:
1. If degree of numerator < degree of denominator:
y=0y=0 is the horizontal asymptote

2. If degree of numerator = degree of denominator:
Divide the coefficients of the highest degree terms from the numerator and denominator to get the horizontal asymptote

3. If degree of numerator > degree of denominator:
There is no horizontal asymptote

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Example: Horizontal Asymptotes

Find the horizontal asymptote(s), if any, of the following functions.
a) f(x)=2x−2x2+3x\displaystyle f(x)=\frac{2x-2}{x^2+3x}

Limit as x→∞‾\underline{x\to\infty}:
lim⁡x→∞ 2x−2x2+3x\displaystyle \lim_{x\to\infty}\ \frac{2x-2}{x^2+3x}
=lim⁡x→∞ 2xx2−2x2x2x2+3xx2\displaystyle =\lim_{x\to\infty}\ \frac{\frac{2x}{\orange{x^2}}-\frac{2}{\orange{x^2}}}{\frac{x^2}{\orange{x^2}}+\frac{3x}{\orange{x^2}}}
=lim⁡x→∞ 2x−2x21+3x\displaystyle =\lim_{x\to\infty}\ \frac{\frac{2}{x}-\frac{2}{x^2}}{1+\frac{3}{x}}
=0−01+0\displaystyle =\frac{0-0}{1+0}
=0=0

Limit as x→−∞‾\underline{x\to-\infty }:
Same as above → lim⁡=0\lim=0

Therefore, the graph has a horizontal asymptote at y=0y=0.

Behaviour near the horizontal asymptote:
f(−1000)=2(−1000)−2(−1000)2+3(−1000)=−2002997000<0f\left(-1000\right)=\frac{2\left(-1000\right)-2}{\left(-1000\right)^2+3\left(-1000\right)}=\frac{-2002}{997000}<0 (below the horizontal asymptote)
f(+1000)=2(+1000)−2(+1000)2+3(−1000)=19989973000>0f\left(+1000\right)=\frac{2\left(+1000\right)-2}{\left(+1000\right)^2+3\left(-1000\right)}=\frac{1998}{9973000}>0 (above the horizontal asymptote)

Short-cut:
The degree of the numerator is 1, the degree of the denominator is 2.
Since degree of numerator < degree of denominator, the graph has one horizontal asymptote y=0y=0.

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b) g(x)=3x3−4x+13−5x3+2x\displaystyle g\left(x\right)=\frac{3x^3-4x+1}{3-5x^3+2x}

Limit as x→∞‾\underline{x\to\infty}:
lim⁡x→∞ 3x3−4x+13−5x3+2x\displaystyle \lim_{x\to\infty}\ \frac{3x^3-4x+1}{3-5x^3+2x}
lim⁡x→∞ 3x3x3−4xx3+1x33x3−5x3x3+2xx3\displaystyle \displaystyle \lim_{x\to\infty}\ \frac{\frac{3x^3}{\orange{x^3}}-\frac{4x}{\orange{x^3}}+\frac{1}{\orange{x^3}}}{\frac{3}{\orange{x^3}}-\frac{5x^3}{\orange{x^3}}+\frac{2x}{\orange{x^3}}}
=lim⁡x→∞ 3−4x2+1x33x3−7+2x2\displaystyle =\lim_{x\to\infty}\ \frac{3-\frac{4}{x^2}+\frac{1}{x^3}}{\frac{3}{x^3}-7+\frac{2}{x^2}}
=3−0+00−5+0\displaystyle =\frac{3-0+0}{0-5+0}
=−35\displaystyle =-\frac{3}{5}

Limit as x→−∞‾\underline{x\to-\infty }:
Same as above → lim⁡=−35\lim=-\frac{3}{5}

Therefore, the graph has a horizontal asymptote at y=−35y=-\frac{3}{5}.

Behaviour near the horizontal asymptote:
f(−1000)=3(−1000)3−4(−1000)+13−5(−1000)3+2(−1000)=−3000000000+4000+13+5000000000−2000<−29999959994999998003>−35f\left(-1000\right)=\frac{3\left(-1000\right)^3-4\left(-1000\right)+1}{3-5\left(-1000\right)^3+2\left(-1000\right)}=\frac{-3000000000+4000+1}{3+5000000000-2000}<\frac{-2999995999}{4999998003}>-\frac{3}{5} (above the horizontal asymptote)
f(+1000)=3(1000)3−4(1000)+13−7(1000)3+2(1000)=2999996001−6999997997>−35f\left(+1000\right)=\frac{3\left(1000\right)^3-4\left(1000\right)+1}{3-7\left(1000\right)^3+2\left(1000\right)}=\frac{2999996001}{-6999997997}>-\frac{3}{5} (above the horizontal asymptote)

Short-cut:
The degree of the numerator is 3, the degree of the denominator is 3.
Since degree of numerator = degree of denominator, the graph has one horizontal asymptote y=3−7 or −37y=\frac{3}{-7}\ or\ -\frac{3}{7}.

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c) h(x)=x2+3x2−x\displaystyle h\left(x\right)=\frac{x^2+3x}{2-x}

Limit as x→∞‾\underline{x\to\infty}:
lim⁡x→∞ x2+3x2−x\displaystyle \lim_{x\to\infty}\ \frac{x^2+3x}{2-x}
=lim⁡x→∞ x2x2+3xx22x2−xx2\displaystyle =\lim_{x\to\infty}\ \frac{\frac{x^2}{\orange{x^2}}+\frac{3x}{\orange{x^2}}}{\frac{2}{\orange{x^2}}-\frac{x}{\orange{x^2}}}
=lim⁡x→∞ 1+3x2x2−1x\displaystyle =\lim_{x\to\infty}\ \frac{1+\frac{3}{x}}{\frac{2}{x^2}-\frac{1}{x}}
=1+00−0\displaystyle =\frac{1+0}{0-0}
=10\displaystyle =\frac{1}{0}
→∞ (DNE)\to\infty\ \left(DNE\right)

Limit as x→−∞‾\underline{x\to-\infty }:
Same as above → lim⁡=DNE\lim=DNE

Therefore, the graph doesn't have any horizontal asymptotes.

Short-cut:
The degree of the numerator is 2, the degree of the denominator is 1.
Since degree of numerator > degree of denominator, the graph doesn't have any horizontal asymptotes.

Practice: Horizontal Asymptotes

Find the horizontal asymptote(s), if any, of the following functions
f(x)=x4−3x3−x\displaystyle f(x)=\frac{x^4-3x}{3-x}

Practice: Vertical & Horizontal Asymptotes

Given the function f(x)=x+12x−1f\left(x\right)=\frac{x+1}{2x-1}, answer the following questions.
Determine the one vertical asymptote of the graph f(x)f\left(x\right).
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Oblique (Slant) Asymptote

An oblique asymptote is a slanted line of the form y=mx+by=mx+b which the graph gets closer and closer to.

When is there an oblique asymptote?
Rational functions where the degree of the numerator is exactly 1 higher than the degree of the denominator has an oblique asymptote.

Steps to finding an oblique asymptote
  1. Divide the numerator by the denominator (using long division or synthetic division)
  2. The quotient is the oblique asymptote

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Example
Find all asymptotes of the graph f(x)=3x2+5x−1x+2f\left(x\right)=\frac{3x^2+5x-1}{x+2}.

Vertical asymptote
The function is undefined when the denominator equals 0:
x+2=0 → x=−2x+2=0\ \to\ x=-2
x→−2−10−f(x)→−∞x→−2+10+f(x)→+∞\begin{array}{|c|c|c|} \hline x\to-2^-&\frac{1}{0^-}&f(x)\to-\infty\\ \hline x\to-2^+&\frac{1}{0^+}&f(x)\to+\infty\\ \hline \end{array}
There, the function has a vertical asymptote at x=−2x=-2.

Oblique asymptote
Since the degree of the numerator is exactly one higher than the denominator, there is an oblique asymptote.
  3x−1x+2)3x2+5x−1‾  3x2+6x____________         −x−1         −x−2____________                    1\begin{array}{cl} &\ \ 3x-1\\ x+2&)\overline{3x^2+5x-1}\\ &\ \ 3x^2+6x\\ &\text{\_\_\_\_\_\_\_\_\_\_\_\_}\\ &\ \ \ \ \ \ \ \ \ -x-1\\ &\ \ \ \ \ \ \ \ \ -x-2\\ &\text{\_\_\_\_\_\_\_\_\_\_\_\_}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \end{array}
So, f(x)=3x2+5x−1x+2=(x+2)(3x−1)+1x+2=(3x−1)+1x+2f\left(x\right)=\frac{3x^2+5x-1}{x+2}=\frac{\left(x+2\right)\left(3x-1\right)+1}{x+2}=\left(3x-1\right)+\frac{1}{x+2}.
As x→±∞x\to\pm\infty, 1x+2→0\frac{1}{x+2}\to0, meaning that the function gets really close to 3x−13x-1.
Therefore, the oblique asymptote is y=3x−1y=3x-1

Horizontal asymptote
lim⁡x→∞ 3x2+5x−1x+2=DNE\displaystyle \lim_{x\to\infty}\ \frac{3x^2+5x-1}{x+2}=DNE
Therefore, there is no horizontal asymptote.


Practice: Asymptotes

Given the following functions, determine which ones have vertical, horizontal, and oblique asymptote(s).

f(x)=x−1x2−4\displaystyle f(x)=\frac{x-1}{x^2-4}

g(x)=2x2x2+1\displaystyle g\left(x\right)=\frac{2x^2}{x^2+1}

h(x)=3x3+x+1x2−1\displaystyle h\left(x\right)=\frac{3x^3+x+1}{x^2-1}
Which function(s) has at least one vertical asymptote?
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Example: Asymptotes

If f(x)=b−ax2−bx\displaystyle f\left(x\right)=\frac{b-ax}{2-bx} has a vertical asymptote at x=5x=5 and a horizontal asymptote at y=1y=1, determine the values of the constants aa and bb.

Vertical asymptote
The function is undefined when the denominator equals 0 and when x=5x=5:
2−b(5)=02-b\left(5\right)=0
2−5b=02-5b=0
2=5b2=5b
b=25b=\frac{2}{5}

Horizontal asymptote
Since we have a rational function where the degree of the numerator and denominator is the same, we can find the horizontal asymptote by dividing the coefficient fo the numerator and denominator:
y=−a−by=\frac{-a}{-b}

Now, we set b=25b=\frac{2}{5} and y=1y=1:
1=−a−251=\frac{-a}{-\frac{2}{5}}
−25=−a-\frac{2}{5}=-a
a=25a=\frac{2}{5}

Therefore, a=25a=\frac{2}{5} and b=25b=\frac{2}{5}.