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Graphs of Rational Functions: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Reading excluded inputs

The figure shows a rational function with exactly the two displayed excluded inputs. Read its real domain from the graph.

The graph of the rational function.

Text description of this figure

A grid with the x-axis from -8 to 4 and the y-axis from -5 to 7, both ticked and labeled at every integer. A dashed vertical line stands at x equals -3. Left of it, a curve comes in from the left edge a little above the x-axis, crosses the x-axis, and falls ever more steeply, running down alongside the dashed line and off the bottom of the frame. Right of the dashed line, a second curve comes down from the top of the frame alongside the dashed line and levels off toward the right edge; it has an open circle at the point x equals -1, y equals 4. Arrows at all four ends of the two branches show them continuing beyond the window. No equation is shown.
1. Hint 1
  Both an open circle and an unbounded break represent missing inputs.
2. Hint 2
  Read their horizontal locations, not their heights.
Answer

All real $x$ except $-3$ and $-1$ .

Full solution

The open circle has horizontal coordinate $-1$ , so that input is missing.
The two branches approach the vertical line $x=-3$ without including any point on that line.
Thus the domain is
$x\ne-3,-1$

The statement guarantees that the figure displays all excluded inputs, so there are no further restrictions to infer.

Answer

All real $x$ except $-3$ and $-1$ .

Key idea

Both holes and vertical asymptotes exclude inputs from a rational function's domain.
Problem 1: Mark complete
Problem 2 A line approached far away

A real function satisfies $f(x)=2+\frac3{x^2+1}$ . Name its horizontal asymptote and explain how the distance to that line behaves for large $|x|$ .
1. Hint 1
  The constant part identifies a candidate line.
2. Hint 2
  The remaining positive fraction can become as small as desired as the denominator grows.
Answer

Horizontal asymptote $y=2$ ; the distance $3/(x^2+1)$ approaches $0$ .

Full solution

The vertical distance to $y=2$ is
$|f(x)-2|=\frac3{x^2+1}$

This positive quantity becomes and stays smaller than any chosen positive distance when $|x|$ is sufficiently large.
It is never zero.
Thus the curve approaches the horizontal line $y=2$ from above.

Answer

Horizontal asymptote $y=2$ ; the distance $3/(x^2+1)$ approaches $0$ .

Key idea

An asymptote describes a line that the curve approaches arbitrarily closely far along its path.
Problem 2: Mark complete
Problem 3 Choosing a coefficient

Find the real number $a$ for which $f(x)=\frac{ax^2+x+1}{2x^2+5}$ has horizontal asymptote $y=3$ .
1. Hint 1
  When the degrees match, the far-out ratio is the ratio of leading coefficients.
2. Hint 2
  The linear numerator term prevents exact division into a constant.
Answer

$a=6$ .

Full solution

For a nonzero $a$ , the two degrees match and the horizontal asymptote is $y=a/2$ .
Thus
$\frac a2=3$ $a=6$

At that value, subtracting $3$ leaves $(x-14)/(2x^2+5)$ , which tends to zero far out but is not the zero expression.
If $a=0$ , the numerator would have lower degree and the asymptote would instead be $0$ .

Answer

$a=6$ .

Key idea

A required horizontal asymptote can determine a leading coefficient.
Problem 3: Mark complete
Problem 4 A missing zero

For $f(x)=\frac{(x-8)^3}{(x-8)^2(2x+3)}$ , identify the domain, every hole and vertical asymptote, and every x-intercept. Explain why an input that zeros the reduced numerator may fail to be an intercept, and decide whether $f$ changes sign across its hole.
1. Hint 1
  Compare how many copies of each denominator factor remain after cancellation.
2. Hint 2
  An intercept must be a point in the original domain.
Answer

Domain $x\ne8,-\frac32$ ; hole $(8,0)$ ; vertical asymptote $x=-\frac32$ ; no x-intercepts. The function changes sign across the hole: negative just left, positive just right.

Full solution

The original denominator excludes $8$ and $-\frac32$ .
Removing two copies of $x-8$ gives
$f(x)=\frac{x-8}{2x+3}$
for allowed inputs.

The reduced denominator is $19$ at $x=8$ , not zero, and the reduced height there is $0$ , giving a hole at $(8,0)$ .
The factor $2x+3$ remains, so $x=-\frac32$ is a vertical asymptote.
The reduced numerator vanishes only at the excluded input $8$ , so no x-intercept exists.

Near $8$ the denominator $2x+3$ is positive, so $f$ has the sign of $x-8$ : negative just left of $8$ and positive just right of it.
The graph passes from below the axis to above it through a point that is missing, without ever touching the axis.

Answer

Domain $x\ne8,-\frac32$ ; hole $(8,0)$ ; vertical asymptote $x=-\frac32$ ; no x-intercepts. The function changes sign across the hole: negative just left, positive just right.

Key idea

A zero of the reduced numerator is an intercept only if the original domain includes that input.
Problem 4: Mark complete
Problem 5 An end-behavior calculation

Find every horizontal or oblique asymptote of $f(x)=\frac{x^3+2x^2+x+4}{x^2+2x}$ , and give a decomposition that explains the end behavior.
1. Hint 1
  Compare degrees to determine which type of line is possible.
2. Hint 2
  Divide the numerator by the denominator and retain the nonzero remainder.
Answer

Oblique asymptote $y=x$ ; no horizontal asymptote; $f(x)=x+\frac{x+4}{x^2+2x}$ for $x\ne0,-2$ .

Full solution

The numerator degree exceeds the denominator degree by one.
Taking $x$ times the denominator and subtracting leaves the remainder $x+4$ , so
$f(x)=x+\frac{x+4}{x^2+2x}$
with $x\ne0,-2$ .

The remainder is nonzero and has lower degree than the denominator, so its quotient tends to $0$ far out.
The graph approaches $y=x$ .
Its growing linear part rules out a horizontal asymptote.
Multiplying the decomposition by the denominator reproduces the original numerator.

Answer

Oblique asymptote $y=x$ ; no horizontal asymptote; $f(x)=x+\frac{x+4}{x^2+2x}$ for $x\ne0,-2$ .

Key idea

When the numerator's degree is one more than the denominator's and the remainder is nonzero, the linear quotient from polynomial division is the oblique asymptote.
Problem 5: Mark complete
Problem 6 When the line is the graph

For a real constant $k$ , let $f_k(x)=\frac{x^2+k}{x^2+13}$ . Using the convention that an asymptote is approached rather than coincided with, find every $k$ for which $f_k$ has no asymptote at all, and describe the graph for those $k$ and for all other $k$ .
1. Hint 1
  Decide first whether the denominator can ever be zero.
2. Hint 2
  Divide: write $f_k(x)$ as $1$ plus a correction, and ask when the correction is the zero expression.
Answer

Only $k=13$ : then $f_{13}(x)=1$ for every real $x$ , and the graph is the whole line $y=1$ , with no hole and no asymptote. For every other $k$ , $y=1$ is a horizontal asymptote the graph never meets, and there is no vertical asymptote.

Full solution

The denominator $x^2+13$ is positive, so every $f_k$ is defined for all real $x$ : there is no hole and no vertical asymptote for any $k$ .

Dividing,
$f_k(x)=1+\frac{k-13}{x^2+13}$

If $k=13$ , the correction is the zero expression and $f_{13}(x)=1$ at every real input.
The graph coincides with the line $y=1$ , so under the stated convention that line is not an asymptote, and there is none.

If $k\ne13$ , the correction is never zero but tends to $0$ as $|x|$ grows, so $y=1$ is a horizontal asymptote that the graph approaches from above when $k>13$ and from below when $k<13$ , without meeting it.

Answer

Only $k=13$ : then $f_{13}(x)=1$ for every real $x$ , and the graph is the whole line $y=1$ , with no hole and no asymptote. For every other $k$ , $y=1$ is a horizontal asymptote the graph never meets, and there is no vertical asymptote.

Key idea

Exact polynomial division makes the function equal to its polynomial quotient throughout its original domain.
Problem 6: Mark complete
Problem 7 Crossing a horizontal line

Let $f(x)=2+\frac{x^2-1}{x^4+1}$ . Determine its horizontal asymptote and every point where it crosses that line. Explain why these crossings do not conflict with the definition of an asymptote.
1. Hint 1
  The added fraction has lower numerator degree than denominator degree.
2. Hint 2
  Its denominator is positive, so the numerator decides both the meeting points and the side of the line.
Answer

Horizontal asymptote $y=2$ ; crossings at $(-1,2)$ and $(1,2)$ ; the crossings do not conflict with the definition.

Full solution

The correction $(x^2-1)/(x^4+1)$ tends to zero for large $|x|$ , so $y=2$ is the horizontal asymptote.
Its denominator is positive everywhere.
The correction vanishes at $-1$ and $1$ .

For $x<-1$ the correction is positive, for $-1<x<1$ it is negative, and for $x>1$ it is positive.
It changes sign at both meetings, so both are crossings.
These finite crossings do not affect the required far-out approach to the line.

Answer

Horizontal asymptote $y=2$ ; crossings at $(-1,2)$ and $(1,2)$ ; the crossings do not conflict with the definition.

Key idea

A rational function may cross its horizontal asymptote before approaching it at large inputs.
Problem 7: Mark complete
Problem 8 A factor that survives

A student says that $F(x)=\frac{x+12}{(x+12)(x^2+15)}$ must have a vertical asymptote, because a denominator factor survives after the common factor cancels. Is the claim true? Give the domain, every hole, and every asymptote.
1. Hint 1
  A vertical asymptote needs an excluded input where the reduced denominator is zero.
2. Hint 2
  Find every real zero of the factor that survives.
Answer

False. The domain is $x\ne-12$ ; there is a hole at $(-12,\frac1{159})$ ; the horizontal asymptote is $y=0$ ; there is no vertical or oblique asymptote.

Full solution

The factor $x^2+15$ is positive, so the original denominator excludes only $-12$ .
Canceling $x+12$ gives
$F(x)=\frac1{x^2+15}$
for $x\ne-12$ .

At $-12$ the reduced denominator is $159$ , not zero, so the graph has a hole at $(-12,\frac1{159})$ .
The surviving factor $x^2+15$ has no real zero, so no input makes the reduced denominator zero, and there is no vertical asymptote.

The numerator has lower degree than the denominator, so the values tend to $0$ far out: the horizontal asymptote is $y=0$ , and there is no oblique asymptote.

Answer

False. The domain is $x\ne-12$ ; there is a hole at $(-12,\frac1{159})$ ; the horizontal asymptote is $y=0$ ; there is no vertical or oblique asymptote.

Key idea

After all common factors cancel, each real zero of the remaining denominator gives a vertical asymptote, and a surviving factor with no real zero gives none.
Problem 8: Mark complete
Problem 9 An added point

A student draws a rational function with vertical asymptote $x=4$ and then adds a solid point $(4,2)$ , saying the graph can cross the asymptote there. Is this a valid graph of the same rational function? Explain.
1. Hint 1
  A vertical asymptote of a rational function occurs at an excluded input.
2. Hint 2
  A solid point asserts an actual function value at its horizontal coordinate.
Answer

No; $(4,2)$ is not a point of the graph of this rational function.

Full solution

The vertical asymptote $x=4$ means the reduced denominator vanishes there, while the numerator does not.
The original rational formula is undefined at that input.

A solid point would instead assert
$f(4)=2$
That would define a different function with an extra assigned value.
It does not create a crossing point for the original rational function.

Answer

No; $(4,2)$ is not a point of the graph of this rational function.

Key idea

An excluded input supplies no point through which the original rational graph could cross a vertical asymptote.
Problem 9: Mark complete
Problem 10 Reading a finite window

The figure shows part of a rational graph close to the horizontal axis near the window edges. A student says this picture alone proves the horizontal asymptote is $y=0$ . Is that conclusion justified without the formula or information about behavior beyond the window? Explain.

Part of the graph of a rational function.

Text description of this figure

A grid with the x-axis from -5 to 5, ticked and labeled at every integer, and the y-axis from 0 to 1.2, ticked and labeled every 0.2. A smooth, symmetric, bell-shaped curve reaches its highest point, height 1, above x equals 0, and falls on both sides. At both edges of the window, x equals -5 and x equals 5, the curve is only a little above the x-axis. The curve simply stops at the window's edges: no arrows on it, no asymptote line and no equation are drawn.
1. Hint 1
  An asymptote describes behavior arbitrarily far out, not only at the displayed endpoints.
2. Hint 2
  Compare the interval covered by the x-axis labels with the inputs the definition of a horizontal asymptote is about.
Answer

No; the finite window alone does not establish a horizontal asymptote.

Full solution

The visible curve being close to $y=0$ at the edges establishes only a finite collection of visual observations.
A horizontal asymptote requires the distance to stay smaller than every positive target distance sufficiently far out.

That stronger statement needs a formula or explicit far-out information.
The plot is suggestive, but it does not by itself establish the claim.

Answer

No; the finite window alone does not establish a horizontal asymptote.

Key idea

A finite graph window suggests end behavior but does not prove an asymptotic statement.
Problem 10: Mark complete

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