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Functions - Asymptotes

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A vertical asymptote occurs at a value x=cx = c where the function f(x)f(x) approaches infinity or negative infinity as xx approaches cc. For rational functions f(x)=P(x)Q(x)f(x) = \frac{P(x)}{Q(x)}, this usually occurs at the roots of the denominator Q(x)Q(x) that are not also roots of the numerator.

Graph showing a vertical asymptote at x = 2 for the function 1/(x-2).
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A horizontal asymptote y=Ly = L describes the behavior of the function as xx approaches positive or negative infinity. For a rational function, if the degree of the numerator equals the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients.

Graph showing a horizontal asymptote at y = 3.
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Exponential functions of the form f(x)=a⋅bx+kf(x) = a \cdot b^{x} + k always possess a horizontal asymptote at y=ky = k. As x→−∞x \to -\infty (for b>1b > 1), the term a⋅bxa \cdot b^x approaches zero, leaving the constant kk.

Exponential graph with horizontal asymptote y = 1.
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Logarithmic functions of the form f(x)=log⁡b(x−h)+kf(x) = \log_{b}(x - h) + k have a vertical asymptote at x=hx = h. This is because the argument of a logarithm must be strictly positive (x−h>0x - h > 0), so the function is undefined for x≤hx \leq h.

📐Formulae

f(x)=ax+bcx+d  ⟹  VA: x=−dc, HA: y=acf(x) = \frac{ax + b}{cx + d} \implies \text{VA: } x = -\frac{d}{c}, \text{ HA: } y = \frac{a}{c}

lim⁡x→cf(x)=±∞  ⟹  x=c is a Vertical Asymptote\lim_{x \to c} f(x) = \pm\infty \implies x = c \text{ is a Vertical Asymptote}

lim⁡x→±∞f(x)=L  ⟹  y=L is a Horizontal Asymptote\lim_{x \to \pm\infty} f(x) = L \implies y = L \text{ is a Horizontal Asymptote}

For f(x)=P(x)Q(x), if deg(P)=deg(Q), HA is y=leading coeff of Pleading coeff of Q\text{For } f(x) = \frac{P(x)}{Q(x)}, \text{ if } \text{deg}(P) = \text{deg}(Q), \text{ HA is } y = \frac{\text{leading coeff of } P}{\text{leading coeff of } Q}

💡Examples

Problem 1:

Find the equations of the asymptotes for the function f(x)=6x−22x+4f(x) = \frac{6x - 2}{2x + 4}.

Solution:

  1. To find the Vertical Asymptote, set the denominator to zero: 2x+4=0  ⟹  2x=−4  ⟹  x=−22x + 4 = 0 \implies 2x = -4 \implies x = -2.
  2. To find the Horizontal Asymptote, look at the ratio of the leading coefficients: y=62  ⟹  y=3y = \frac{6}{2} \implies y = 3.

Explanation:

The vertical asymptote occurs where the function is undefined (denominator is zero). The horizontal asymptote is found by evaluating the limit as xx approaches infinity, which for linear-over-linear functions is the ratio of the xx coefficients.

Problem 2:

Determine the horizontal asymptote of g(x)=5−2ex−1g(x) = 5 - 2e^{x-1}.

Solution:

As x→−∞x \to -\infty, the term ex−1→0e^{x-1} \to 0. Therefore, g(x)→5−2(0)=5g(x) \to 5 - 2(0) = 5. The horizontal asymptote is y=5y = 5.

Explanation:

For exponential functions y=a⋅bx−h+ky = a \cdot b^{x-h} + k, the horizontal asymptote is always y=ky = k because the exponential part approaches zero in one direction of xx.

Problem 3:

Find the vertical asymptote of the function h(x)=ln⁡(3x+9)−4h(x) = \ln(3x + 9) - 4.

Solution:

The argument of a logarithm must be greater than zero. The vertical asymptote occurs where the argument equals zero: 3x+9=0  ⟹  3x=−9  ⟹  x=−33x + 9 = 0 \implies 3x = -9 \implies x = -3.

Explanation:

Logarithmic functions are undefined for values that make the inner argument zero or negative; the boundary of this domain is the vertical asymptote.

Problem 4:

Identify the equations of the vertical and horizontal asymptotes for the function f(x)=4x2−1x2−9f(x) = \frac{4x^2 - 1}{x^2 - 9}.

Graph of (4x^2-1)/(x^2-9) showing asymptotes at x=-3, x=3, and y=4.

Solution:

  1. Vertical Asymptotes: Set the denominator to zero: x2−9=0  ⟹  (x−3)(x+3)=0x^2 - 9 = 0 \implies (x-3)(x+3) = 0. Thus, x=3x = 3 and x=−3x = -3 are vertical asymptotes.
  2. Horizontal Asymptote: Compare the degrees of the numerator and denominator. Both are degree 2. The horizontal asymptote is the ratio of the leading coefficients: y=41=4y = \frac{4}{1} = 4.

Explanation:

Vertical asymptotes occur where the denominator is zero. Horizontal asymptotes are found by looking at the limit as xx goes to infinity, which for equal-degree polynomials is the ratio of coefficients.

Problem 5:

Find the asymptote of the function f(x)=3x+2−5f(x) = 3^{x+2} - 5.

Graph of y = 3^(x+2) - 5 showing the horizontal asymptote at y = -5.

Solution:

The function is an exponential function of the form y=a⋅bx−h+ky = a \cdot b^{x-h} + k.

  1. As x→−∞x \to -\infty, 3x+2→03^{x+2} \to 0.
  2. Therefore, f(x)→0−5=−5f(x) \to 0 - 5 = -5.
  3. The horizontal asymptote is y=−5y = -5.

Explanation:

Exponential functions have a single horizontal asymptote. The vertical shift of the parent function y=3xy=3^x determines the position of this asymptote.