3.4 Sketching and interpreting rational functions
A rational graph is assembled from domain restrictions, discontinuities, intercepts, sign intervals, and end behaviour. Each feature constrains the final sketch.
3.4.1 A systematic graphing process
First factor and state the domain. Record holes before cancelling. Find intercepts, vertical asymptotes, and one-sided behaviour. Use zeros and asymptotes to divide the number line into sign intervals. Determine horizontal behaviour from degrees, then sketch branches that respect every feature.
Worked Example: Graphing a rational function with two asymptotes
For \(f(x)=(2x-1)/(x+2)\), the domain excludes \(-2\), the \(x\)-intercept is \(1/2\), the \(y\)-intercept is \(-1/2\), the vertical asymptote is \(x=-2\), and the horizontal asymptote is \(y=2\). Test points determine the sign of each branch.
Check Your Work
Domain \(x\ne1\); \(x\)-intercept \(-3\); \(y\)-intercept \(-3\); vertical asymptote \(x=1\); horizontal asymptote \(y=1\).Worked Example: Graphing with a hole
For \(r(x)=(x^2-4)/(2x^2+2x)\),
\[r(x)=\frac{(x-2)(x+2)}{2x(x+1)}.\]
No factor cancels, so there is no hole. The vertical asymptotes are \(x=0,-1\), zeros are \(x=\pm2\), and the horizontal asymptote is \(y=1/2\).
Check Your Work
Yes. It simplifies from \((x-1)(x+1)/[(x+1)(x+2)]\) and has a hole at \(x=-1\).3.4.2 Domain and range from a rational graph
Domain is read horizontally and excludes holes and vertical asymptotes. Range is read vertically and may exclude a horizontal value or the output at a hole. Solving \(y=f(x)\) for \(x\) can help identify excluded output values when the graph is difficult to read.
Worked Example: Finding the range of a reciprocal transformation
For \(f(x)=3+2/(x-4)\), the vertical asymptote is \(x=4\) and the horizontal asymptote is \(y=3\). Since \(2/(x-4)\) can never be zero, the range is \((-\infty,3)\cup(3,\infty)\).
Check Your Work
Domain excludes \(-2\); range excludes \(-1\).Worked Example: Checking a physically restricted branch
If \(H(q)=40q/(q+5)\) models head for \(q\ge0\), the algebraic vertical asymptote \(q=-5\) lies outside the physical domain. On \(q\ge0\), \(H\) begins at 0, increases, and approaches 40 from below.
Check Your Work
It begins at 0, increases for nonnegative time, and approaches 10 without reaching it.Applied Problem: Analyzing a saturating treatment model. Removal efficiency is modelled by \(E(C)=90C/(C+4)\) for dosage \(C\ge0\). Find the intercept, identify the relevant asymptotes, describe the physical range, and calculate \(E(8)\).
Check Your Work
The intercept is \((0,0)\). The algebraic vertical asymptote \(C=-4\) is outside the physical domain. The horizontal asymptote is \(E=90\), so the physical range is \([0,90)\). At \(C=8\), \(E=90(8)/12=60\).3.4.3 Practice Problems
- List the intercepts and asymptotes of \((x-2)/(x+1)\).
Check Your Work
\(x\)-intercept 2; \(y\)-intercept \(-2\); vertical asymptote \(x=-1\); horizontal asymptote \(y=1\). - Determine the sign of \((x-3)/(x+2)\) on \((-2,3)\).
Check Your Work
Negative. - Find the domain and range of \(2+1/(x-5)\).
Check Your Work
Domain excludes 5; range excludes 2. - Identify all graphing features of \(x/(x^2-4)\).
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Domain excludes \(\pm2\); intercept \((0,0)\); vertical asymptotes \(x=\pm2\); horizontal asymptote \(y=0\). - Explain why a graph cannot pass through a hole.
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The corresponding input is excluded from the original domain. - A dosage response is \(E(C)=75C/(C+3)\) for \(C\ge0\). Find \(E(6)\) and the limiting efficiency.
Check Your Work
\(E(6)=50\); the limiting efficiency is 75. - A hydraulic index is \(H(q)=20q/(q+10)\) for \(q\ge0\). State its physical range.
Check Your Work
\([0,20)\). - A model \(P(q)=(q-5)/(q+1)\) is used only for \(q\ge0\). Which algebraic features lie outside the physical domain?
Check Your Work
The vertical asymptote \(q=-1\) lies outside the physical domain.