Leading Term In A Polynomial

Section i.6: Polynomials and Rational Functions

Polynomial Functions

Terminology of Polynomial Functions

A polynomial is a function that can be written as \[ f(10)=a_0+a_1 x+a_2 ten^2+\dots+a_n x^n \]

Each of the \(a_i\) constants are called coefficients and tin can exist positive, negative, or zero, and be whole numbers, decimals, or fractions.

A term of the polynomial is whatever one piece of the sum, that is any \( a_i x^i \). Each individual term is a transformed power function.

The degree of the polynomial is the highest power of the variable that occurs in the polynomial.

The leading term is the term containing the highest power of the variable: the term with the highest degree.

The leading coefficient is the coefficient of the leading term.

Considering of the definition of the "leading" term we ofttimes rearrange polynomials so that the powers are descending: \[ f(x)=a_n 10^n+a_{n-1}10^{n-1}\dots a_2 x^2+a_1 x+a_0 \]

Example 1

Place the degree, leading term, and leading coefficient of the polynomial \( f(x)=3+2x^2-4x^3 \)

The caste is 3, the highest power on \(x\). The leading term is the term containing that power, \( -4x^3 \). The leading coefficient is the coefficient of that term, -4.

Brusque Run Behavior: Intercepts

As with whatsoever function, the vertical intercept can be found past evaluating the function at an input of zero. Since this is evaluation, it is relatively easy to do it for a polynomial of any degree. To find horizontal intercepts, we need to solve for when the output will be zippo. For general polynomials, this can be a challenging prospect. Consequently, we volition limit ourselves to three cases:

The polynomial tin can exist factored using known methods: greatest common cistron and trinomial factoring.
The polynomial is given in factored grade.
Technology is used to make up one's mind the intercepts.

Instance 2

Find the horizontal intercepts of \( f(x)=10^six-3x^4+2x^two \).

We can attempt to factor this polynomial to find solutions for \(f(10) = 0\):

\(x^half-dozen-3x^4+2x^ii=0\)
\(ten^ii(x^4-3x^2+2)=0\)	Factoring out the greatest common factor
\(ten^ii(x^2-1)(x^two-2)=0\)	Factoring the within as a quadratic in \(x^two\)

Then break apart to notice solutions:

\( x^ii=0 \)		\( x^2-one=0 \)		\( ten^two-2=0 \)
\(x=0\)		\(10^2=1\)		\( x^2=2 \)
\(x=0\)	or	\( x=\pm ane \)	or	\( x=\pm\sqrt{ii} \)

This gives us five horizontal intercepts.

Example iii

Notice the horizontal intercepts of \( h(t)=t^3+4t^2+t-6 \)

Since this polynomial is not in factored course, has no common factors, and does non appear to exist factorable using techniques nosotros know, nosotros can plow to engineering science to observe the intercepts.

Graphing this role, it appears there are horizontal intercepts at \(t =\) -3, -2, and 1.

Nosotros could check these are right past plugging in these values for \(t\) and verifying that \( h(-three)=h(-2)=h(1)=0 \).

Solving Polynomial Inequalities

One application of our power to discover intercepts and sketch a graph of polynomials is the ability to solve polynomial inequalities. It is a very mutual question to enquire when a role will be positive and negative, and one we volition employ afterwards in this course.

Case four

Solve \( (x+three)(10+one)^2(x-iv)\gt 0 \)

As with all inequalities, we kickoff by solving the equality \( (x+three)(x+1)^2(x-iv)= 0 \), which has solutions at \(ten =\) -3, -1, and iv. We know the function can only change from positive to negative at these values, so these divide the inputs into 4 intervals.

We could choose a exam value in each interval and evaluate the office \( f(x)=(x+iii)(10+1)^two(x-4) \) at each test value to determine if the office is positive or negative in that interval:

Interval	Test \( x \) in interval	\( f(\text{exam value}) \)	\( \gt 0 \) or \( \lt 0 \)?
\( ten\lt -iii \)	-4	72	\( \gt 0 \)
\( -three\lt x\lt -ane \)	-2	-vi	\( \lt 0 \)
\( -one \lt 10 \lt iv \)	0	-12	\( \lt 0 \)
\( x\gt 4 \)	v	288	\( \gt 0 \)

On a number line this would look like:

From our test values, nosotros tin determine this part is positive when \(x \lt -3 \) or \(x \gt four\), or in interval notation, \( (-\infty,-3) \cup (4,\infty) \).

Rational Functions

Rational functions are the ratios, or fractions, of polynomials. They can arise from both simple and circuitous situations.

Example five

You plan to drive 100 miles. Find a formula for the time the trip will accept as a part of the speed you lot bulldoze.

You lot may recollect that multiplying speed by fourth dimension will give you distance. If we let \(t\) stand for the drive fourth dimension in hours, and \(v\) correspond the velocity (speed or rate) at which we drive, then \(vt=\)altitude. Since our distance is fixed at 100 miles, \(vt=100\). Solving this relationship for the time gives usa the part we desired: \[t(five)=\frac{100}{5}\]

Notice that this is a transformation of the reciprocal toolkit function, \( f(10)=\dfrac{1}{x} \). Several natural phenomena, such every bit gravitational force and volume of sound, behave in a manner inversely proportional to the square of another quantity. For example, the volume, \(V\), of a audio heard at a distance \(d\) from the source would exist related past \( Five=\dfrac{one thousand}{d^2} \) for some constant value \(k\). These functions are transformations of the reciprocal squared toolkit function \( f(ten)=\dfrac{ane}{10^two} \).

We accept seen the graphs of the basic reciprocal office and the squared reciprocal function from our review of toolkit functions. These graphs take several important features.

reciprocal squared — \( f(x)=\frac{1}{ten^ii} \)

Let's begin by looking at the reciprocal function, \( f(10)=\dfrac{i}{ten} \). As you well know, dividing past zero is not allowed and therefore zip is not in the domain, and so the function is undefined at an input of nix.

Short Run beliefs:

As the input values approach nix from the left side (taking on very small, negative values), the function values get very large in the negative direction (in other words, they approach negative infinity). We write: \( x\to 0^- \), \( f(x)\to -\infty \).

Every bit we arroyo nothing from the right side (small, positive input values), the function values become very large in the positive direction (approaching infinity). Nosotros write: as \( x\to 0^+ \), \( f(x)\to \infty \).

This behavior creates a vertical asymptote. An asymptote is a line that the graph approaches. In this case the graph is approaching the vertical line \(x = 0\) every bit the input becomes close to zero.

Long Run behavior:

As the values of 10 approach infinity, the function values approach 0. Also, equally the values of 10 approach negative infinity, the function values arroyo 0. Symbolically: equally \( x\to\pm\infty \), \( f(x)\to 0 \).

Based on this long run beliefs and the graph nosotros tin run across that the function approaches 0 but never actually reaches 0, it just "levels off" as the inputs become large. This behavior creates a horizontal asymptote. In this case the graph is approaching the horizontal line \( f(ten)=0 \) as the input becomes very large in the negative and positive directions.

Vertical and Horizontal Asymptotes

A vertical asymptote of a graph is a vertical line \(x = a\) where the graph tends towards positive or negative infinity equally the inputs arroyo \(a\). As \( 10\to a \), \( f(10)\to\pm\infty \).

A horizontal asymptote of a graph is a horizontal line \( y=b \) where the graph approaches the line as the inputs go large. As \( x\to\pm\infty \), \( f(x)\to b \).

Instance vi

Sketch a graph of the reciprocal function shifted 2 units to the left and upward three units. Identify the horizontal and vertical asymptotes of the graph, if whatever.

Transforming the graph left 2 and up 3 would consequence in the function \( f(ten)=\dfrac{i}{x+2}+3 \), or equivalently, past giving the terms a mutual denominator, \[f(x)=\dfrac{3x+7}{x+2}.\]

Shifting the toolkit role would give us this graph. Detect that this equation is undefined at \(x = -two\), and the graph also is showing a vertical asymptote at \(x = -ii\). As \( 10\to -two^- \), \( f(x)\to -\infty \), and as \( x\to -2^+ \), \( f(10)\to \infty \).

Every bit the inputs grow large, the graph appears to exist leveling off at output values of 3, indicating a horizontal asymptote at \( y=three\). As \( x\to\pm\infty \), \( f(x)\to three \)

Notice that horizontal and vertical asymptotes go shifted left two and up iii forth with the function.

A general rational function is the ratio of whatever two polynomials.

Rational Function

A rational function is a role that tin can exist written equally the ratio of 2 polynomials, \(P(10)\) and \(Q(10)\). \[f(x)=\frac{P(10)}{Q(ten)}=\frac{a_0+a_1 x+a_2 ten^two+\dots+a_p x^p}{b_0+b_1 x+b_2 x^two+\dots+b_q ten^q}\]

Rational functions can ascend from real situations.

Instance vii

A large mixing tank currently contains 100 gallons of water, into which 5 pounds of sugar have been mixed. A tap volition open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a charge per unit of ane pound per minute. Discover the concentration (pounds per gallon) of carbohydrate in the tank after \(t\) minutes.

Find that the amount of water in the tank is changing linearly, equally is the amount of sugar in the tank. We can write an equation independently for each:\[\text{h2o}=100+10t \qquad \text{saccharide}=v+1t\]

The concentration, \(C\), will be the ratio of pounds of sugar to gallons of water: \[C(t)=\frac{5+t}{100+10t}\]

Vertical Asymptotes of Rational Functions

The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zilch.

Horizontal Asymptote of Rational Functions

The horizontal asymptote of a rational role can be determined by looking at the degrees of the numerator and denominator.

Degree of denominator > degree of numerator: Horizontal asymptote at \( y=0 \).
Caste of denominator < degree of numerator: No horizontal asymptote.
Degree of denominator = degree of numerator: Horizontal asymptote at ratio of leading coefficients, \( y=\dfrac{a_p}{b_q} \) (\(p\) and \(q\) are equal in this case).

Example 8

In the sugar concentration problem from earlier, we created the equation \( C(t)=\frac{v+t}{100+10t} \). Find the horizontal asymptote and interpret it in context of the scenario.

Both the numerator and denominator are linear (degree 1), so since the degrees are equal, there volition be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is t, with coefficient i. In the denominator, the leading term is \(10t\), with coefficient 10. The horizontal asymptote will be at the ratio of these values: As \( t \to \infty \), \( C(t)\to \frac{1}{10} \). This office will have a horizontal asymptote at \( y=\frac{one}{x} \).

This tells us that equally the input gets large, the output values will approach \( \frac{1}{10} \). In context, this means that as more time goes by, the concentration of sugar in the tank volition arroyo one tenth of a pound of sugar per gallon of water or \( \frac{1}{x} \) pounds per gallon.

Case ix

Observe the horizontal and vertical asymptotes of the function \[f(x)=\frac{(x-2)(ten+three)}{(x-1)(10+2)(x-5)}.\]

Offset, note this part has no inputs that brand both the numerator and denominator zero, so there are no potential holes. The function volition have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at \(x =\) ane, -ii, and five, indicating vertical asymptotes at these values.

The numerator has degree two, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will abound faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and and then as \( x\to\pm\infty \), \( f(x)\to 0 \). This office volition have a horizontal asymptote at \( y=0 \).

Every bit with all functions, a rational function will have a vertical intercept when the input is zero, if the function is defined at zero. It is possible for a rational function to not take a vertical intercept if the function is undefined at zero.

Likewise, a rational function will have horizontal intercepts at the inputs that cause the output to be naught (unless that input corresponds to a hole). It is possible there are no horizontal intercepts. Since a fraction is only equal to zero when the numerator is goose egg, horizontal intercepts volition occur when the numerator of the rational role is equal to zero.

Instance 10

Notice the intercepts of \[f(x)=\frac{(x-ii)(10+3)}{(x-1)(x+2)(x-five)}.\]

We can notice the vertical intercept by evaluating the function at zero: \[f(0)=\frac{(0-2)(0+3)}{(0-1)(0+2)(0-5)}=\frac{-6}{x}=-\frac{3}{five}.\]

The horizontal intercepts will occur when the function is equal to zero: \[ \begin{align*} 0=& \frac{(ten-2)(x+3)}{(x-one)(x+2)(ten-five)} \qquad \text{(This is zero when the numerator is zero.)}\\ 0=& (x-two)(x+3)\\ x=& 2, -3. \stop{align*} \]