True or False? The numerator of a rational function plays no part in determining its domain.

True. The domain is restricted only by the denominator, since the function is undefined exactly where it would require division by zero. The numerator can be any polynomial at all without changing which input values are allowed.

What does the value of a rational function measure at each input?

It measures the relative size of the numerator polynomial compared with the denominator polynomial at that input value. That comparison is only defined for input values that lie in the function's domain.

What determines the end behavior of a rational function?

How the degrees of the numerator and denominator compare. For input values of large magnitude each polynomial is dominated by its leading term, so the end behavior follows the quotient of the leading terms .

True or False? A rational function equals zero wherever its denominator equals zero.

False. Where the denominator is zero the function is undefined , so it has no value there at all. A fraction takes the value zero only when its numerator does.

Define vertical asymptote for a rational function.

A vertical asymptote is a vertical line at an input value where the function is undefined, but which the graph approaches more and more closely from either side. It is a boundary for the graph rather than a part of it, and is usually drawn as a dashed line.

True or False? Every input value that makes the denominator zero gives a vertical asymptote.

False. A vertical asymptote can only occur where the denominator is zero, but not every such value produces one. When the numerator is zero there as well, with at least equal multiplicity, the graph has a hole instead.

Define hole for a rational function.

A hole is a point at an input value where the function is undefined, but where the function does not show unbounded behavior on either side. It is often drawn as an open circle on the graph.

What is a quick way to tell a hole from a vertical asymptote?

Simplify the expression by cancelling common factors, then look at the denominator that is left. If it is no longer zero at that input value there is a hole , and if it is still zero there is a vertical asymptote .

How does a function's behavior near a hole differ from its behavior near a vertical asymptote?

Near a hole the output values converge towards a single finite value from both sides. Near a vertical asymptote they increase or decrease without bound instead.

In what order are the terms of a polynomial long division answer found?

One term at a time, working downwards in powers of the variable, starting from the highest power. Each term is multiplied by the divisor and the result subtracted, exactly as in ordinary long division of numbers.

What are three uses of polynomial long division?

Factoring a polynomial, simplifying an algebraic fraction, and finding the equation of a slant asymptote. All three follow from splitting a polynomial into a factor pair, with or without a remainder term.

What is almost always the first step when asked about the graph of a rational function?

Factor the numerator and the denominator separately . Once both are factored, the common factors, the remaining denominator zeros and the remaining numerator zeros can all be read off together.

How do you factor or expand a rational function?

Treat the numerator and the denominator separately , using the same techniques as for any polynomial. To factor, take out common factors and factor quadratics; to expand, multiply out each one in turn.

True or False? A value that gives a hole is still excluded from the domain of a rational function.

True. The denominator is zero at that value, so the function is undefined there whether the graph shows a hole or a vertical asymptote. Cancelling the common factor changes the expression, but it does not put the missing input value back.

AP®MathsCollege BoardPrecalculusFlashcardsPolynomial & Rational FunctionsRational Functions

Rational Functions (College Board AP® Precalculus): Flashcards

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FrontRational Functions
Define rational function.

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Cards in this collection (51)

Define rational function.
A rational function is one that can be written as a quotient of two polynomial functions, $r (x) = \frac{p (x)}{q (x)}$ .
Here $p (x)$ is the polynomial in the numerator and $q (x)$ is the polynomial in the denominator.
What is the domain of a rational function, unless a question says otherwise?
All real numbers $x$ for which the denominator is not zero.
The function is undefined wherever the denominator equals zero, because division by zero is not defined.
How do you find the domain of $r (x) = \frac{x + 5}{x^{2} - 9}$ ?
Set the denominator equal to zero, solve, and exclude those values.
Here $x^{2} - 9 = 0$ gives $x = 3$ and $x = - 3$ , so the domain is all real numbers with $x \neq 3$ and $x \neq - 3$ .
The function $r$ is given by $r (x) = \frac{x^{2} + 3}{x - 1}$ . Fill in the two gaps to evaluate $r (4)$ .
$r (4) = \frac{16 + 3}{______- 1} = \frac{19}{______}$
The completed working is:
$r (4) = \frac{16 + 3}{4 - 1} = \frac{19}{3}$
Substitute the input value into the numerator and the denominator separately, then form the quotient.
True or False?
The numerator of a rational function plays no part in determining its domain.
True.
The domain is restricted only by the denominator, since the function is undefined exactly where it would require division by zero.
The numerator can be any polynomial at all without changing which input values are allowed.
How can a graphing calculator solve $r (x) = k$ for a rational function $r$ ?
Graph $y = r (x)$ and the horizontal line $y = k$ on the same axes, then find where they intersect.
The $x$ coordinates of those intersections are the solutions, and it is worth zooming out far enough to be sure none has been missed.
What does the value of a rational function measure at each input?
It measures the relative size of the numerator polynomial compared with the denominator polynomial at that input value.
That comparison is only defined for input values that lie in the function's domain.
What determines the end behavior of a rational function?
How the degrees of the numerator and denominator compare.
For input values of large magnitude each polynomial is dominated by its leading term, so the end behavior follows the quotient of the leading terms.
The denominator of a rational function has a higher degree than the numerator. What is the end behavior?
The denominator dominates, so the function approaches zero and there is a horizontal asymptote at $y = 0$ .
For $r (x) = \frac{x + 4}{x^{3} - 2 x}$ the quotient of leading terms is $\frac{x}{x^{3}} = \frac{1}{x^{2}}$ , which tends to zero.
The numerator and denominator of a rational function have equal degree. What is the end behavior?
Neither dominates, so there is a horizontal asymptote at $y = \frac{a_{n}}{b_{n}}$ , the ratio of the two leading coefficients.
For $r (x) = \frac{3 x^{2} + 1}{5 x^{2} - 2}$ the horizontal asymptote is $y = \frac{3}{5}$ .
True or False?
In $\frac{x^{3} - 7 x - 43}{3 - x}$ the leading term of the denominator is $3$ .
False.
The leading term is the term with the highest power of $x$ , not the term written first, so it is $- x$ .
The quotient of leading terms is therefore $\frac{x^{3}}{- x} = - x^{2}$ .
The numerator of a rational function has a higher degree than the denominator. What is the end behavior?
The numerator dominates, so there is no horizontal asymptote and the function increases or decreases without bound.
Its end behavior matches the polynomial given by the quotient of the leading terms, so $\frac{8 x^{5} + x - 2}{2 x^{2} - 5 x + 1}$ behaves like $4 x^{3}$ .
A rational function $r$ has a horizontal asymptote at $y = b$ . Fill in the two gaps in its end-behavior limits.
$\lim_{x \to \infty} r (x) =______and \lim_{x \to______} r (x) = b$
The completed statement is:
$\lim_{x \to \infty} r (x) = b and \lim_{x \to - \infty} r (x) = b$
The output values get arbitrarily close to $b$ and stay close as the input values increase or decrease without bound.
When does the graph of a rational function have a slant asymptote?
When the degree of the numerator exceeds the degree of the denominator by exactly $1$ .
The function still increases or decreases without bound, but its graph gets closer and closer to a slanted line as the input values grow.
True or False?
A rational function can have end behavior identical to that of a quadratic.
True.
When the numerator's degree exceeds the denominator's by $2$ , the quotient of the leading terms is a quadratic and the end behavior matches it.
For $\frac{6 x^{4} + 1}{3 x^{2} - 2}$ that quotient is $2 x^{2}$ , so the function increases without bound in both directions.
Where do the real zeros of $r (x) = \frac{p (x)}{q (x)}$ occur?
At the values of $x$ where $p (x) = 0$ but $q (x) \neq 0$ .
A zero of the numerator that is also a zero of the denominator is excluded from the domain, so it cannot be a zero of $r$ .
The function $h$ is given by $h (x) = \frac{(x + 4) (x - 3)}{(x + 1) (x - 3)}$ . What are its real zeros?
The only real zero is $x = - 4$ .
The numerator is zero at $x = - 4$ and at $x = 3$ , but the denominator is also zero at $x = 3$ , so that value is excluded.
True or False?
A rational function equals zero wherever its denominator equals zero.
False.
Where the denominator is zero the function is undefined, so it has no value there at all.
A fraction takes the value zero only when its numerator does.
Where can the graph of a rational function touch or cross the $x$ -axis?
It can meet the $x$ -axis only at its real zeros.
Those are the only input values at which the function takes the value zero, so the graph cannot reach the axis anywhere else.
Which values of $x$ divide a rational function into intervals of constant sign?
The real zeros of both the numerator and the denominator.
Within each interval between consecutive such values the function is either positive everywhere or negative everywhere, so those are the only points at which it can change sign.
How do you find the intervals where $r (x) > 0$ for a rational function?
Find the real zeros of the numerator and of the denominator, then test a single value of $x$ inside each resulting interval.
Only the sign of each test value matters, so you can usually decide it without working out the exact value.
A test value gives $f (- 4) = \frac{(- 1) (- 5) (- 9)}{(- 3) (- 5) (49)}$ . Fill in the two gaps.
The numerator is $______$ and the denominator is positive, so $f (- 4)$ is $______$ overall.
The completed sentence is:
The numerator is negative and the denominator is positive, so $f (- 4)$ is negative overall.
Three negative factors multiply to give a negative, and the exact value is never needed.
Define vertical asymptote for a rational function.
A vertical asymptote is a vertical line at an input value where the function is undefined, but which the graph approaches more and more closely from either side.
It is a boundary for the graph rather than a part of it, and is usually drawn as a dashed line.
When does $r (x) = \frac{p (x)}{q (x)}$ have a vertical asymptote at $x = a$ ?
When $q (a) = 0$ and $p (a) \neq 0$ , so the denominator vanishes there but the numerator does not.
For example, $\frac{x + 4}{x - 3}$ has a vertical asymptote at $x = 3$ .
Both $p$ and $q$ are zero at $x = a$ . When is there still a vertical asymptote there?
When the multiplicity of $a$ as a zero of the denominator is greater than its multiplicity as a zero of the numerator.
So $\frac{(x + 4) {(x - 3)}^{5}}{{(x - 3)}^{6}}$ has a vertical asymptote at $x = 3$ since $6 > 5$ , and otherwise there would be a hole instead.
True or False?
Every input value that makes the denominator zero gives a vertical asymptote.
False.
A vertical asymptote can only occur where the denominator is zero, but not every such value produces one.
When the numerator is zero there as well, with at least equal multiplicity, the graph has a hole instead.
What do $\lim_{x \to a^{+}}$ and $\lim_{x \to a^{-}}$ mean?
The first is the limit as $x$ approaches $a$ from above, that is from the right, using input values greater than $a$ .
The second is the limit as $x$ approaches $a$ from below, from the left, using input values less than $a$ .
What happens to the values of a rational function near a vertical asymptote?
They increase or decrease without bound, so each one-sided limit there is $\infty$ or $- \infty$ .
The denominator becomes arbitrarily close to zero while the numerator does not, which is what makes the quotient grow without limit.
How do you decide whether a one-sided limit at a vertical asymptote is $\infty$ or $- \infty$ ?
Evaluate the function at a value close to the asymptote on that side and look at the sign.
For $s (x) = \frac{x + 4}{x - 3}$ at $x = 2$ the numerator is positive and the denominator negative, so $\lim_{x \to 3^{-}} s (x) = - \infty$ .
For $s (x) = \frac{x + 4}{x - 3}$ , approaching $x = 3$ from above keeps both the numerator and the denominator positive. Fill in the gap.
$\lim_{x \to 3^{+}} s (x) =______$
The completed statement is:
$\lim_{x \to 3^{+}} s (x) = \infty$
A positive numerator over a positive denominator that is shrinking towards zero grows without bound.
Define hole for a rational function.
A hole is a point at an input value where the function is undefined, but where the function does not show unbounded behavior on either side.
It is often drawn as an open circle on the graph.
Both $p$ and $q$ are zero at $x = c$ . When does $r (x) = \frac{p (x)}{q (x)}$ have a hole there?
When the multiplicity of $c$ as a zero of the numerator is greater than or equal to its multiplicity as a zero of the denominator.
So $\frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}}$ has a hole at $x = 3$ , where the two multiplicities are equal.
What is a quick way to tell a hole from a vertical asymptote?
Simplify the expression by cancelling common factors, then look at the denominator that is left.
If it is no longer zero at that input value there is a hole, and if it is still zero there is a vertical asymptote.
How does a function's behavior near a hole differ from its behavior near a vertical asymptote?
Near a hole the output values converge towards a single finite value from both sides.
Near a vertical asymptote they increase or decrease without bound instead.
How do you find the $y$ coordinate of a hole?
Simplify the expression algebraically, then substitute the input value into the simplified form.
For $\frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}}$ this gives $\frac{x + 4}{x + 1}$ , which at $x = 3$ is $\frac{7}{4}$ , so the hole is at $(3, \frac{7}{4})$ .
Consider $\frac{(x + 5) (x + 7)}{(x - 3) (x + 5)}$ and fill in the two gaps.
The common factor $(x + 5)$ cancels, and the simplified denominator $(x - 3)$ is $______$ zero at $x = - 5$ , so the graph has a $______$ at that input value.
The completed sentence is:
The common factor $(x + 5)$ cancels, and the simplified denominator $(x - 3)$ is not zero at $x = - 5$ , so the graph has a hole at that input value.
Had the simplified denominator still been zero there, it would have been a vertical asymptote instead.
How is the behavior of a rational function near a hole written in limit notation?
If the output values become arbitrarily close to $L$ as $x$ approaches $c$ , then $\lim_{x \to c} r (x) = L$ .
The limits from the left and from the right both equal $L$ , which is why a single two-sided limit can be written.
True or False?
$f (x) = \frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}}$ and $g (x) = \frac{x + 4}{x + 1}$ are the same function.
False.
The expressions simplify to one another, but the two functions have different domains and so are not equal.
At $x = 3$ the function $g$ takes the value $\frac{7}{4}$ , while $f$ is undefined and has a hole there.
Complete the identity produced when a polynomial $f$ is divided by a polynomial $g$ , where $q$ is the quotient and $r$ is the remainder.
$f (x) = g (x)______(x) +______(x)$
The completed identity is:
$f (x) = g (x) q (x) + r (x)$
The degree of the remainder $r$ is always less than the degree of the divisor $g$ .
In what order are the terms of a polynomial long division answer found?
One term at a time, working downwards in powers of the variable, starting from the highest power.
Each term is multiplied by the divisor and the result subtracted, exactly as in ordinary long division of numbers.
When do you stop the process of polynomial long division?
When what is left has a lower degree than the polynomial you are dividing by.
Whatever remains at that point is the remainder, so dividing by a degree $1$ divisor means continuing until only a constant is left.
True or False?
If polynomial long division leaves a remainder of zero, the divisor is a factor of the polynomial.
True.
With $r (x) = 0$ the identity becomes $f (x) = g (x) q (x)$ , which is exactly what it means for $g$ to be a factor.
Dividing $x^{3} + 6 x^{2} - 9 x - 14$ by $x - 2$ leaves no remainder, giving $(x - 2) (x^{2} + 8 x + 7)$ .
What are three uses of polynomial long division?
Factoring a polynomial, simplifying an algebraic fraction, and finding the equation of a slant asymptote.
All three follow from splitting a polynomial into a factor pair, with or without a remainder term.
How does polynomial long division give the equation of a slant asymptote?
Dividing gives $\frac{f (x)}{g (x)} = q (x) + \frac{r (x)}{g (x)}$ , and that remainder term tends to zero as the input values grow.
So the function gets closer and closer to the quotient, making $y = q (x)$ the slant asymptote.
Long division gives $3 x^{3} - 4 x^{2} - 15 x + 7 = (x^{2} - 5) (3 x - 4) - 13$ . Fill in the two gaps for $f (x) = \frac{3 x^{3} - 4 x^{2} - 15 x + 7}{x^{2} - 5}$ .
Rewriting gives $f (x) = 3 x - 4 - \frac{13}{x^{2} - 5}$ , and that fraction tends to $______$ as $x$ grows, so the slant asymptote is the line $y =______$ exactly.
The completed sentence is:
Rewriting gives $f (x) = 3 x - 4 - \frac{13}{x^{2} - 5}$ , and that fraction tends to 0 as $x$ grows, so the slant asymptote is the line $y = 3 x - 4$ exactly.
The remainder term always vanishes in the end behavior, because the divisor has the higher degree.
What is almost always the first step when asked about the graph of a rational function?
Factor the numerator and the denominator separately.
Once both are factored, the common factors, the remaining denominator zeros and the remaining numerator zeros can all be read off together.
Once a rational function is fully factored, what do a common factor, a remaining denominator zero and a remaining numerator zero each give?
A common factor gives a hole, a remaining denominator zero gives a vertical asymptote, and a remaining numerator zero gives an $x$ -intercept.
Many questions about rational functions can be answered from that single comparison of factors.
Fill in the two gaps about which form of a rational function reveals what.
The $______$ form of the numerator and denominator shows the zeros, holes and vertical asymptotes, while the $______$ form shows the degrees and so settles the end behavior.
The completed sentence is:
The factored form of the numerator and denominator shows the zeros, holes and vertical asymptotes, while the standard form shows the degrees and so settles the end behavior.
Being able to convert between them means either kind of information is always available.
How do you factor or expand a rational function?
Treat the numerator and the denominator separately, using the same techniques as for any polynomial.
To factor, take out common factors and factor quadratics; to expand, multiply out each one in turn.
Which of $\frac{x^{2} - 7 x + 10}{x^{2} - x - 2}$ and $\frac{x - 5}{x^{2} - x - 2}$ has a hole, and where?
The first one has a hole at $x = 2$ , because it factors as $\frac{(x - 5) (x - 2)}{(x - 2) (x + 1)}$ with $(x - 2)$ common to both.
The second has no common factor, so $x = 2$ gives a vertical asymptote there instead.
True or False?
A value that gives a hole is still excluded from the domain of a rational function.
True.
The denominator is zero at that value, so the function is undefined there whether the graph shows a hole or a vertical asymptote.
Cancelling the common factor changes the expression, but it does not put the missing input value back.

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