Other Function Families

Rational functions: asymptotes and holes

A rational function is a fraction whose numerator and denominator are polynomials. To understand its graph, factor the numerator and denominator first.

• A zero of the denominator that remains after common factors are canceled creates a vertical asymptote at that $x$-value.
• A factor that cancels creates a hole at that $x$-value instead.
• A horizontal asymptote comes from comparing the degrees of the numerator and denominator. If the degrees are equal, the asymptote is the ratio of the leading coefficients. If the numerator's degree is smaller, the asymptote is $y = 0$. If the numerator's degree is larger, there is no horizontal asymptote at all: the function grows without bound instead of leveling off.

Finding asymptotes

Worked example. Find the horizontal asymptote of $f(x) = \dfrac{4x + 1}{2x - 3}$.

Compare the degrees of the numerator and denominator. Both have degree 1, so the asymptote is the ratio of leading coefficients.

$y = \dfrac{4}{2} = 2$

For the vertical asymptote of $\dfrac{x+4}{x-6}$, set the denominator equal to zero: $x = 6$. Do not report the zero of the numerator ($x = -4$); that is where the function crosses the $x$-axis. A common trap with horizontal asymptotes is to report the numerator's constant or the function's $y$-intercept. For $\dfrac{5}{x+3}$ the numerator's degree (0) is smaller than the denominator's (1), so the asymptote is $y = 0$, not $y = 5$.

For a hole, cancel and then plug the hole's $x$-value into the simplified expression. The function $f(x) = \dfrac{x^2 - 9}{x - 3} = \dfrac{(x-3)(x+3)}{x-3}$ simplifies to $x + 3$ for $x \ne 3$. There is a hole at $x = 3$, and its $y$-value is $3 + 3 = 6$. The hole's $y$-value is not $0$, and it is not "undefined": the graph simply skips that single point at height $6$.

Logarithms

A logarithm answers one question: what exponent produces this result? The expression $\log_a(b)$ asks what power $a$ must be raised to in order to get $b$. Evaluating a logarithm directly, without a calculator, means recognizing $b$ as a power of $a$.

• $\log_3(81) = 4$ because $3^4 = 81$.
• $\log_2(64) = 6$ because $2^6 = 64$.
• $\log_5(125) = 3$ because $5^3 = 125$.

The answer is the exponent, not the base and not the number inside. If you are unsure, build up the powers: $3, 9, 27, 81$ is four multiplications' worth of 3s. Count the factors carefully, since stopping one short is the usual mistake.

Exponential growth and decay

Exponential models take the form

$$A(t) = A_0(1 + r)^t,$$

where $A_0$ is the starting amount and $r$ is the growth or decay rate written as a decimal, positive for growth and negative for decay. A decay of $15\%$ per period uses the factor $1 - 0.15 = 0.85$; growth of $5\%$ uses $1.05$. Evaluate the exponent first, and then multiply by the starting amount. For $A(t) = 200(0.85)^t$, $A(4) = 200(0.85)^4 \approx 104.40$.

Be careful with the number of periods. Using an exponent of 5 instead of 4 is a very common off-by-one error. Also resist the shortcut of multiplying rate by time: $1000(1.05)^3 \approx 1157.63$, while the "simple interest" value $1000(1 + 0.05\cdot 3) = 1150$ is a tempting wrong choice.

Geometric sequences

A geometric sequence follows the same structure in discrete steps: each term is the previous term multiplied by a constant ratio $r$, so $a_n = a_1 r^{\,n-1}$.

Worked example. A geometric sequence has $a_1 = 4$ and $r = 3$. Find $a_5$.

Each term multiplies the previous term by $r$, applied $(n-1)$ times from the first term.

$a_5 = 4\cdot 3^4 = 4\cdot 81 = 324$

Just as with arithmetic sequences, the number of multiplications is one less than the term number. To reach $a_6$ from $a_1$, multiply by $r$ five times.

Rational equations

Equations that contain fractions with the variable in the denominator need an extra step: check for extraneous solutions.

Worked example. Solve $\dfrac{x + 1}{x - 3} = \dfrac{5}{x - 3} + 2$.

Recognize. Both sides share the same denominator, $x - 3$. Clear it by multiplying every term by that denominator, then check the result for extraneous solutions.

Set up. $(x + 1) = 5 + 2(x - 3)$

Solve. $x + 1 = 5 + 2x - 6$, so $x + 1 = 2x - 1$, and $2 = x$.

Check. $x = 2$ does not make the original denominator zero ($2 - 3 = -1 \ne 0$), so the solution is valid.

Best route. Hand algebra clears the fractions directly. Desmos can graph both sides to confirm the intersection, but it does not set up the multiplication step.

Common error. Finding a solution but forgetting to check whether it makes the original denominator zero, which would make it extraneous.

Remember to multiply every term, including the plain number $2$, by the common denominator.

Where this appears as a model

An investment growing at a fixed percentage rate is the classic exponential model. If $1,000 is invested at 5% annual interest, it grows to $1000(1.05)^{10} \approx \$1{,}628.89$ after 10 years.

Strategy summary

• Factor first. Surviving denominator zeros are vertical asymptotes; canceled factors are holes.
• Horizontal asymptote: compare degrees (equal means ratio of leading coefficients, smaller on top means $y=0$, larger on top means none).
• A logarithm is an exponent. Ask "what power?" and count the factors.
• In $A_0(1+r)^t$, use $1-r$ for decay and count periods exactly.
• After clearing denominators, always check that your answer does not make an original denominator zero.