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Amy Ferguson Moncure

Calculus BC ยท Unit 1

Asymptotes

A leftover zero in the denominator is a vertical asymptote. Matching degrees give a horizontal asymptote from the leading coefficients.

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Key Ideas Easy Medium Hard

VA: leftover zero in the denominator

HA: \(\lim_{x \to \infty} f(x)\)

What to Remember

Vertical asymptote

After canceling, a factor of the denominator is still 0 and the numerator is not. Then \(x = a\) is a vertical asymptote.

Horizontal asymptote

Look at \(\lim_{x \to \infty} f(x)\) and \(\lim_{x \to -\infty} f(x)\). For a rational, compare degrees. Same degree: ratio of leading coefficients.

Exponential decay

\(e^{-kt} \to 0\) as \(t \to \infty\) when \(k > 0\). A leftover constant is the horizontal asymptote.

A hole is not a VA

If the numerator and denominator share the factor, cancel first. A canceled factor is a hole, not a vertical asymptote.

Level 1

Easy

Name a vertical or horizontal asymptote from a rational.

Example
Find the vertical asymptote of \(y = \dfrac{x + 1}{(x - 5)^2}\).

The denominator is 0 at \(x = 5\).

The numerator is 6, which is not 0.

\(x = 5\)

Answer: \(x = 5\)

1
Find the horizontal asymptote of \(y = \dfrac{4x^2 - 1}{2x^2 + 7}\).
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The degrees are equal.

Ratio of leading coefficients: \(\dfrac{4}{2} = 2\).

\(y = 2\)

Answer: \(y = 2\)

2
Find the horizontal asymptote of \(y = \dfrac{3x}{x^2 + 1}\).
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The degree of the bottom is larger.

\(y = 0\)

Answer: \(y = 0\)

3
Find \(\lim_{t \to \infty}\bigl(12 + 5e^{-0.4t}\bigr)\).
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\(e^{-0.4t} \to 0\)

\(12 + 5 \cdot 0\)

\(12\)

Answer: \(12\)

4
Find the vertical asymptote of \(y = \dfrac{x - 3}{x + 8}\).
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\(x = -8\)

Answer: \(x = -8\)

5
Find the horizontal asymptote of \(y = \dfrac{6x^2 + 1}{3x^2 - 5}\).
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\(\dfrac{6}{3} = 2\)

\(y = 2\)

Answer: \(y = 2\)

Level 2

Medium

Cancel first, then decide hole vs asymptote, or find a limit at infinity.

Example
Let \(f(x) = \dfrac{x^2 - 4}{x^2 - x - 2}\). Find the vertical asymptote and any hole.

\(x^2 - 4 = (x - 2)(x + 2)\)

\(x^2 - x - 2 = (x - 2)(x + 1)\)

\(f(x) = \dfrac{x + 2}{x + 1}\) for \(x \neq 2\).

Hole at \(x = 2\).

Vertical asymptote \(x = -1\).

Answer: hole at \(x = 2\), VA \(x = -1\)

1
Find \(\lim_{x \to \infty} \dfrac{5x^2 - 3}{x^2 + 8}\).
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Divide top and bottom by \(x^2\), or use leading coefficients.

\(\dfrac{5}{1}\)

\(5\)

Answer: \(5\)

2
Find \(\lim_{t \to \infty}\bigl(9 + 4e^{-0.2(t - 6)}\bigr)\).
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\(e^{-0.2(t - 6)} \to 0\)

\(9\)

Answer: \(9\)

3
Find the horizontal asymptote of \(y = \dfrac{-2x^2 + x}{x^2 - 9}\).
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\(y = -2\)

Answer: \(y = -2\)

4
Let \(f(x) = \dfrac{3x^2 - 12}{x^2 - 16}\). Identify the hole and the vertical asymptote.
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\(\dfrac{3(x - 2)(x + 2)}{(x - 4)(x + 4)}\) does not cancel 2 with 4.

No hole from canceling.

Vertical asymptotes at \(x = 4\) and \(x = -4\).

Answer: VA at \(x = \pm 4\), no hole

5
Write an equation for the horizontal asymptote of \(f(x) = \dfrac{-5x^2 + 2x}{x^2 + 4}\).
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Degrees equal. Leading coefficients \(-5\) over 1.

\(y = -5\)

Answer: \(y = -5\)

Level 3

Hard

Unknown constants in a rational, or a limit at infinity from a piecewise model.

Example
Let \(f(x) = \dfrac{bx - 3x^2}{x^2 + px + q}\). The graph has a vertical asymptote at \(x = 1\) and a removable discontinuity at \(x = -4\). Show that \(p = 3\) and \(q = -4\), then find \(b\).

The denominator is 0 at the VA and at the hole, so \(x^2 + px + q = (x - 1)(x + 4) = x^2 + 3x - 4\).

\(p = 3\), \(q = -4\).

A hole at \(x = -4\) means the numerator is also 0 there.

\(b(-4) - 3(16) = 0\)

\(-4b - 48 = 0\)

\(-4b = 48\)

\(b = -12\)

Answer: \(b = -12\)

1
Let \(f(x) = \dfrac{-12x - 3x^2}{x^2 + 3x - 4}\). What value of \(f(-4)\) makes \(f\) continuous at \(-4\)?
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Factor.

\(\dfrac{-3x(x + 4)}{(x - 1)(x + 4)}\)

Cancel \(x + 4\).

\(\dfrac{-3x}{x - 1}\)

Substitute \(x = -4\).

\(\dfrac{-3(-4)}{-4 - 1}\)

\(\dfrac{12}{-5}\)

\(-\dfrac{12}{5}\)

Answer: \(-\dfrac{12}{5}\)

2
Write an equation for the horizontal asymptote of \(f(x) = \dfrac{-12x - 3x^2}{x^2 + 3x - 4}\).
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Degrees of top and bottom are both 2. Leading coefficients \(-3\) over 1.

\(y = -3\)

Answer: \(y = -3\)

3
A mass is modeled by \(Y(t) = 16 + 5e^{-0.25(t - 11)}\) for \(t > 11\). Find \(\lim_{t \to \infty} Y(t)\) and explain what it means.
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\(e^{-0.25(t - 11)} \to 0\)

\(Y(t) \to 16\)

In the long run the mass approaches 16.

Answer: \(16\)

4
Let \(f(x) = \dfrac{cx - 2x^2}{x^2 - 5x + 6}\). If there is a hole at \(x = 2\), find \(c\).
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\(x^2 - 5x + 6 = (x - 2)(x - 3)\)

Numerator 0 at 2: \(2c - 8 = 0\).

\(c = 4\)

Answer: \(4\)

5
Using \(c = 4\) in that function, is \(x = 3\) a hole or a vertical asymptote?
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Numerator at 3: \(4(3) - 2(9) = 12 - 18 = -6 \neq 0\).

Vertical asymptote.

Answer: vertical asymptote

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