Rational functions show up in real-world situations involving rates, concentrations, and ratios. A rational function is simply one polynomial divided by another—it’s a fraction where both the top and bottom are polynomials.
Common applications include:
Mixture problems (concentration of solutions)
Rate problems (speed, work rates)
Average cost in business
Population density
The key is setting up the ratio correctly and understanding what happens as time passes or quantities change.
Understanding Asymptotes in Context
The Main Idea
Asymptotes tell us about the long-term behavior of rational functions—what happens as values get very large or approach restricted values.Horizontal Asymptote: A horizontal line [latex]y = c[/latex] that shows what value the function approaches as [latex]x[/latex] gets very large (positively or negatively). This represents the long-run behavior or end behavior.Vertical Asymptote: A vertical line [latex]x = a[/latex] where the function is undefined (the denominator equals zero). The graph “bends around” this line, shooting up or down toward infinity.
Recall: To find a horizontal asymptote, look at the leading terms of the numerator and denominator. If both have the same degree, divide their coefficients. For vertical asymptotes, set the denominator equal to zero and solve.
A large mixing tank currently contains [latex]100[/latex] gallons of water into which [latex]5[/latex] pounds of sugar have been mixed. A tap opens, pouring [latex]10[/latex] gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of [latex]1[/latex] pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after [latex]12[/latex] minutes. Is that a greater concentration than at the beginning?
Let [latex]t[/latex] = number of minutes since the tap opened.
Set up equations for water and sugar:
Water: [latex]W(t) = 100 + 10t[/latex] gallons
Sugar: [latex]S(t) = 5 + t[/latex] pounds
Concentration is the ratio:
[latex]C(t) = \frac{S(t)}{W(t)} = \frac{5 + t}{100 + 10t}[/latex]
After 12 minutes:
[latex]\begin{align} C(12) &= \frac{5 + 12}{100 + 10(12)} \ &= \frac{17}{220} \ &\approx 0.077 \text{ pounds per gallon} \end{align}[/latex]
At the beginning:
[latex]\begin{align} C(0) &= \frac{5 + 0}{100 + 10(0)} \ &= \frac{5}{100} \ &= \frac{1}{20} = 0.05 \text{ pounds per gallon} \end{align}[/latex]
Since [latex]0.077 > 0.05[/latex], the concentration is greater after 12 minutes.
Long-term behavior:
The horizontal asymptote is [latex]y = \frac{1}{10} = 0.1[/latex] (divide leading coefficients). This means the concentration will approach [latex]0.1[/latex] pounds per gallon in the long run.
When working with mixture problems, always identify what’s being added and at what rate. The concentration is always amount of substance divided by total volume.
A rational inequality is like a rational equation, but instead of an equal sign, it uses an inequality symbol ([latex]<[/latex], [latex]\leq[/latex], [latex]>[/latex], or [latex]\geq[/latex]). These inequalities ask: “For which values of [latex]x[/latex] is this rational expression positive (or negative)?”
The key difference from polynomial inequalities: we must be extra careful about values that make the denominator zero—these create vertical asymptotes and are always excluded from solutions.
Solving Rational Inequalities
Rewrite the inequality so one side is zero and the other side is a single rational expression.
Identify critical values:
Zeros of the numerator (where the expression equals zero)
Zeros of the denominator (where the expression is undefined—these are ALWAYS excluded)
Plot critical values on a number line to create intervals.
Test a value from each interval to determine if the expression is positive or negative there.
Select intervals that satisfy the inequality.
Write your answer in interval notation:
Use parentheses [latex]( )[/latex] for excluded values
Use brackets [latex][ ][/latex] only for numerator zeros when the inequality includes “or equal to”
NEVER use brackets for denominator zeros—they’re always excluded
These divide the number line into three intervals:
[latex]x < -2[/latex]
[latex]-2 < x < 4[/latex]
[latex]x > 4[/latex]
Test each interval
Interval
Test Point
Sign of [latex]\frac{x - 4}{x + 2}[/latex]
[latex]x < -2[/latex]
[latex]x = -3[/latex]
[latex]\frac{-7}{-1} = 7[/latex] → positive ✓
[latex]-2 < x < 4[/latex]
[latex]x = 0[/latex]
[latex]\frac{-4}{2} = -2[/latex] → negative
[latex]x > 4[/latex]
[latex]x = 5[/latex]
[latex]\frac{1}{7}[/latex] → positive ✓
Write the solution
We want where the expression is greater than 0 (positive), so we select the intervals where our tests were positive.
Solution: [latex](-\infty, -2) \cup (4, \infty)[/latex]
Note: We use parentheses at [latex]-2[/latex] because it makes the denominator zero (undefined), and at [latex]4[/latex] because the inequality is strictly greater than zero (not “or equal to”).
Denominator zeros are ALWAYS excluded from the solution—they make the expression undefined. Even if the inequality includes “or equal to,” you cannot include values that make the denominator zero.
Solve [latex]\frac{x + 3}{x - 5} \leq 0[/latex]
Identify critical values
Numerator: [latex]x + 3 = 0 \Rightarrow x = -3[/latex]
Write the solution
We want [latex]\leq 0[/latex], which means negative or zero. The expression is negative on [latex](-3, 5)[/latex] and equals zero at [latex]x = -3[/latex].
Solution: [latex][-3, 5)[/latex]
Note: We use a bracket at [latex]-3[/latex] because the inequality includes “or equal to” and [latex]-3[/latex] makes the numerator zero (which gives 0). We use a parenthesis at [latex]5[/latex] because it makes the denominator zero (always excluded).
Create a sign chart! Draw a number line, mark your critical values, test each interval, and label each region as positive or negative. This visual representation makes it much easier to see which intervals satisfy your inequality.