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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: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra I. You can skip it.

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Problem 1 of 10

Problem 1 Where the denominator vanishes

Find the domain of $f(x)=\frac{4x-3}{x^2+5x-14}$ .
1. Hint 1
  A rational function's domain excludes only the inputs that make its denominator zero.
2. Hint 2
  Write the quadratic denominator as two linear factors, then set each factor to zero.
Answer

Domain: every real number except $-7$ and $2$ .

Full solution

Factor the denominator:
$x^2+5x-14=(x+7)(x-2)$

This product is zero at $x=-7$ and at $x=2$ , so those are the two excluded inputs.

A numerator never restricts a domain, so no other input is excluded.

Answer

Domain: every real number except $-7$ and $2$ .

Key idea

The zeros of a rational function's denominator are exactly the inputs its domain excludes.
Problem 1: Mark complete
Problem 2 The two kinds of gap

For $f(x)=\frac{x^2-3x-10}{x^2-x-20}$ , name every input the domain excludes and say for each whether the graph has a hole or a vertical asymptote there. Give the height of any hole.
1. Hint 1
  An excluded input gives a hole only when every copy of its factor cancels away.
2. Hint 2
  Factor the top and the bottom, and cancel what you can.
Answer

Excluded: $5$ and $-4$ . Hole at $\left(5,\frac{7}{9}\right)$ ; vertical asymptote $x=-4$ .

Full solution

The denominator factors as $(x-5)(x+4)$ , so the domain excludes $5$ and $-4$ .
The numerator factors as $(x-5)(x+2)$ .

The factor $x-5$ cancels, leaving $\frac{x+2}{x+4}$ at every input except $5$ and $-4$ .
No copy of $x-5$ is left below, so $5$ gives a hole, while the copy of $x+4$ that is left makes $x=-4$ a vertical asymptote.

The height of the hole is the value of the reduced form there,
$\frac{5+2}{5+4}=\frac{7}{9}$

Answer

Excluded: $5$ and $-4$ . Hole at $\left(5,\frac{7}{9}\right)$ ; vertical asymptote $x=-4$ .

Key idea

Cancel as far as it goes: a factor that disappears leaves a hole at the reduced height, and a copy still below leaves a vertical asymptote.
Problem 2: Mark complete
Problem 3 The numerator record

A rational function has a numerator of degree $5$ and a denominator with leading term $3x^5$ . Its horizontal asymptote is $y=-2$ . Find the numerator leading term.
1. Hint 1
  Equal degrees make the horizontal asymptote the ratio of the leading coefficients.
2. Hint 2
  Only the two leading terms matter far out; write the equation the ratio of leading coefficients must satisfy.
Answer

$-6x^5$ .

Full solution

If the numerator leading coefficient is $a$ , equal degrees give
$\frac a3=-2$
Multiplying by $3$ gives $a=-6$ .

The degree is stated to be $5$ , so the leading term is $-6x^5$ .
Its ratio to $3x^5$ has the required height $-2$ .

Answer

$-6x^5$ .

Key idea

A horizontal asymptote and a known denominator leading term can determine an unknown numerator leading term.
Problem 3: Mark complete
Problem 4 The unfinished branches

The figure marks two points of $y=-2/x$ . Complete the graph in the window, then give the coordinates of its point with $x=-4$ and of its point with $y=-4$ , and explain why no point of the graph lies on either axis.

Two points of $y = -2/x$ , on axes ruled in half units.

Text description of this figure

Coordinate axes on a square grid, with x and y each running from negative four to four and a number at every whole unit. Short extra ticks mark every half unit along both axes, so a height of one half can be read off. Two points are marked with filled dots and labeled with their coordinates: the point (1, negative 2) below the x-axis on the right, and the point (negative 1, 2) above the x-axis on the left. Nothing else is drawn, and no curve passes through the two points.
1. Hint 1
  Each allowed input has one output, and dividing $-2$ by a positive input gives a negative result.
2. Hint 2
  Plot a few more pairs, such as the outputs at $2$ and at $-2$ , before joining anything; the input $0$ is not allowed.
3. Hint 3
  Think about what would make the output zero, and whether that can happen here.
Answer

Two branches, in the second and fourth quadrants, approaching both axes; the points are $\left(-4,\frac{1}{2}\right)$ and $\left(\frac{1}{2},-4\right)$ ; the graph meets neither axis.

Full solution

A positive input gives a negative output and a negative input gives a positive output, so the graph lies in the fourth and second quadrants.
Besides the two marked points, $(2,-1)$ and $(-2,1)$ are on it.

As the size of $x$ grows, the output shrinks toward $0$ ; as $x$ nears $0$ , the output grows without bound in size.
So each branch flattens toward the x-axis at one end and races along the y-axis at the other, and the two branches stay apart because the input $0$ is missing.

At $x=-4$ the output is $-2/(-4)$ , so one requested point is $\left(-4,\frac{1}{2}\right)$ .
An output of $-4$ needs $-2/x=-4$ , that is $x=\frac{1}{2}$ , so the other requested point is $\left(\frac{1}{2},-4\right)$ .

No point of the graph has $x=0$ , because that input is excluded.
No point has $y=0$ either, because a fraction is zero only where its numerator is zero, and this numerator is the constant $-2$ .

The completed graph of $y = -2/x$ , with the two given points and the two requested points marked.

Answer

Two branches, in the second and fourth quadrants, approaching both axes; the points are $\left(-4,\frac{1}{2}\right)$ and $\left(\frac{1}{2},-4\right)$ ; the graph meets neither axis.

Key idea

An excluded input splits the domain in two, so a reciprocal graph comes in two separate branches, and a constant nonzero numerator keeps it off both axes.
Problem 4: Mark complete
Problem 5 The guide lines

A function has the form $f(x)=a/(x-h)+k$ , with $a\ne0$ . Its graph in the figure has the shown asymptotes and marked point. Find its rule and the exact x-intercept.

The graph of $f$ with its two asymptotes and one marked point.

Text description of this figure

A coordinate grid numbered at every whole unit, with x running from negative six to four and y from negative three to eight. A dashed vertical line labeled x equals negative 2 and a dashed horizontal line labeled y equals 3 cross inside the window. A curve labeled f has two separate branches. The left branch lies above the dashed horizontal line: it comes down from the top edge between negative 3 and negative 2, and flattens toward the dashed horizontal line as it runs left, reaching a height of 4 at the left edge. The right branch lies below the dashed horizontal line: it rises steeply from the bottom edge between negative 2 and negative 1, crosses the x-axis between negative 1 and 0, passes through the marked point (0, 1), and flattens toward the dashed horizontal line on its way to the right edge, reaching a height of about 2 and one third at the right edge. A filled dot labeled (0, 1) marks that point on the y-axis.
1. Hint 1
  The asymptotes give $h$ and $k$ .
2. Hint 2
  Substitute the marked point to find $a$ , then solve for output zero.
Answer

$f(x)=-\frac4{x+2}+3$ ; x-intercept $(-\frac23,0)$ .

Full solution

The asymptotes are $x=-2$ and $y=3$ , so $h=-2$ and $k=3$ .
The point $(0,1)$ gives
$1=\frac a2+3$
Thus $a=-4$ .

For the x-intercept, the denominator must be nonzero.
Setting the output to zero gives
$\frac4{x+2}=3$
Multiplication by $x+2\ne0$ gives $4=3x+6$ , hence $x=-2/3$ .

This input is allowed, and substitution makes the output zero.
The rule also returns $1$ at input $0$ , checking the given point.

Answer

$f(x)=-\frac4{x+2}+3$ ; x-intercept $(-\frac23,0)$ .

Key idea

The asymptotes locate a shifted reciprocal, while one point sets its scale.
Problem 5: Mark complete
Problem 6 The faster ride

A cycle path is $48$ km long, so riding it at a steady speed of $x$ km per hour takes $\frac{48}{x}$ hours. State the one speed the rule $\frac{48}{x}$ excludes, and say what the ride time does as the speed drops toward it. Riding $4$ km per hour faster would save $1$ hour: find the slower speed.
1. Hint 1
  A fixed distance divided by a smaller and smaller speed gives a larger and larger time.
2. Hint 2
  Write the two ride times, set their difference equal to $1$ hour, and multiply through by both denominators.
Answer

The rule excludes $x=0$ , and the time grows without bound as the speed drops toward $0$ ; the slower speed is $12$ km per hour.

Full solution

The rule $\frac{48}{x}$ has no value at $x=0$ , the one speed it excludes.
As the speed drops toward $0$ the time grows without bound: $24$ hours at $2$ km per hour, $48$ hours at $1$ km per hour, and $96$ hours at $\frac{1}{2}$ km per hour.

The faster ride takes $\frac{48}{x+4}$ hours, so saving one hour means
$\frac{48}{x}-\frac{48}{x+4}=1$

A speed is positive, so neither denominator is zero and multiplying every term by $x(x+4)$ is safe.
That leaves
$48(x+4)-48x=x(x+4)$

The left side is $192$ , so the quadratic is $x^2+4x-192=0$ , which factors as
$(x-12)(x+16)=0.$

The candidate $-16$ is not a speed, so the slower speed is $12$ km per hour.
Checking, the ride takes $\frac{48}{12}=4$ hours at that speed and $\frac{48}{16}=3$ hours at $16$ km per hour, one hour less.

Answer

The rule excludes $x=0$ , and the time grows without bound as the speed drops toward $0$ ; the slower speed is $12$ km per hour.

Key idea

A fixed distance over a varying speed is a rational rule that bars the speed $0$ , and a candidate from clearing its denominators counts only if the situation allows it.
Problem 6: Mark complete
Problem 7 Clearing the denominators

Solve $\frac{x}{x-7}-\frac{7}{x+7}=\frac{98}{x^2-49}$ , checking every candidate in the original equation.
1. Hint 1
  The third denominator is a difference of squares, and two inputs are barred before any work is done.
2. Hint 2
  Multiply every term by $(x-7)(x+7)$ , then compare what the resulting equation gives with those barred inputs.
Answer

No solution: the candidates are $7$ and $-7$ , and the original equation excludes both.

Full solution

Since $x^2-49=(x-7)(x+7)$ , the three denominators bar $7$ and $-7$ from the start.

At every other input, multiplying each term by $(x-7)(x+7)$ leaves
$x(x+7)-7(x-7)=98$

The left side is $x^2+7x-7x+49$ , so the equation becomes $x^2=49$ , with candidates $7$ and $-7$ .

Those two candidates are exactly the barred inputs, so neither solves the original equation and it has no solution.
A sample allowed input confirms it: at $x=9$ the two sides are $\frac{65}{16}$ and $\frac{49}{16}$ , which differ.

Answer

No solution: the candidates are $7$ and $-7$ , and the original equation excludes both.

Key idea

Clearing denominators can hand back only the inputs the original fractions bar, and the equation then has no solution at all.
Problem 7: Mark complete
Problem 8 Bo's comparison

For $f(x)=\frac{x^2+x}{x^2+1}$ , Bo says the graph never reaches its horizontal asymptote. Find that horizontal asymptote, determine every point, if any, at which the graph meets it, and then decide whether Bo is right.
1. Hint 1
  The degrees and leading coefficients identify the horizontal asymptote.
2. Hint 2
  Set the function equal to that height and check whether the resulting input is allowed.
Answer

Bo is wrong; the asymptote is $y=1$ , and the graph meets it at $(1,1)$ .

Full solution

The degrees are equal and the leading coefficients are both $1$ , so the horizontal asymptote is $y=1$ .
The denominator is positive for every real input.

Equality with the asymptote requires
$x^2+x=x^2+1$
Hence $x=1$ .
Substitution gives $f(1)=2/2=1$ , so the point is $(1,1)$ .

A horizontal asymptote describes distant behavior and need not be avoided at ordinary inputs.

Answer

Bo is wrong; the asymptote is $y=1$ , and the graph meets it at $(1,1)$ .

Key idea

A rational graph may meet its horizontal asymptote even though its far-out behavior approaches that same height.
Problem 8: Mark complete
Problem 9 The paired factors

Let $P,Q$ be nonzero polynomials, and suppose $P/Q$ has a horizontal asymptote. For a real number $a$ , Ana says multiplying both its numerator and denominator by $x-a$ leaves that horizontal asymptote unchanged. Is she correct? Explain.
1. Hint 1
  For every input where both expressions are defined, the added factors cancel.
2. Hint 2
  Compare degrees and leading coefficients before and after the multiplication.
Answer

Yes; the horizontal asymptote stays unchanged.

Full solution

The added factor increases both degrees by $1$ and multiplies both leading coefficients by $1$ .
Therefore the degree comparison is unchanged.

If the numerator degree was smaller, it remains smaller and the asymptote stays $y=0$ .
If the degrees were equal, they remain equal and the same leading-coefficient ratio sets the asymptote.

The new expression excludes $x=a$ if it was previously allowed, but that single finite input does not change the distant behavior.

Answer

Yes; the horizontal asymptote stays unchanged.

Key idea

Adding the same nonzero polynomial factor above and below preserves the degree comparison and leading-coefficient ratio that set a horizontal asymptote.
Problem 9: Mark complete
Problem 10 The missing numerator

Find a quadratic $p(x)$ in standard form so that $f(x)=p(x)/(x^2-2x-8)$ has a hole at $(4,2)$ and a vertical asymptote at $x=-2$ . The leading coefficient of $p$ is $1$ .
1. Hint 1
  A hole at input $4$ requires a factor $x-4$ in the numerator.
2. Hint 2
  Write $p(x)=(x-4)(x+c)$ and use the stated height after cancellation.
Answer

$p(x)=x^2+4x-32$ .

Full solution

The original denominator factors as $(x-4)(x+2)$ and excludes both $4$ and $-2$ .
A quadratic with leading coefficient $1$ canceling $x-4$ has the form $(x-4)(x+c)$ .

The reduced value at $4$ must be $2$ , so
$\frac{4+c}{6}=2$
giving $c=8$ .
Expanding yields
$p(x)=x^2+4x-32$

The reduced function is $(x+8)/(x+2)$ , away from the exclusions.
At $4$ its value would be $2$ , so the original has the required hole.
At $-2$ the numerator is $6\ne0$ , leaving the vertical asymptote.

Answer

$p(x)=x^2+4x-32$ .

Key idea

Prescribed holes and asymptotes constrain which factors the numerator must cancel and which it must leave behind.
Problem 10: Mark complete

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