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Here are a few recommended readings before getting started with this lesson.
A rational function is a function that contains a rational expression. Any function that can be written as the quotient of two polynomial functions $p(x)$ and $q(x)$ is a rational function.
$f(x)=q(x)p(x) ,q(x) =0$
The graph of a rational function can be a smooth continuous curve, or it can have jumps, breaks, or holes. By looking at the graph of a function, functions can be categorized into two groups: continuous or discontinuous.
A point of discontinuity of a function is a point with the $x-$coordinate that makes the function value undefined. A point of discontinuity can also be considered as an excluded value of the function rule.
Write as a power
$a_{2}−b_{2}=(a+b)(a−b)$
Simplify quotient
holein the graph.
When a function cannot be redefined so that the point of discontinuity becomes a valid input, it is called a non-removable discontinuity. Consider, for example, $y=x+11 .$ Since $x=-1$ is not in the domain of the function, there is a point of discontinuity at $x=-1.$
This is a non-removable discontinuity because there is no way to redefine the function so that it becomes continuous at $x=-1.$Total Annual Income | Population |
---|---|
$I(t)=0.02t+15000t+100000 $ | $P(t)=0.36t+10800$ |
Substitute expressions
$ba /c=b⋅ca $
Multiply parentheses
Add terms
$C(t)$ is undefined | |
---|---|
$0.02t+1=0$ | $0.36t+10800=0$ |
$t=-50$ | $t=-30000$ |
On the way to the pencil factory, LaShay noticed that road maintenance work is being carried out to fill a hole
on the road.
Factor out $3x$
Factor out $x_{2}$
Factor out $(x+1)$
Notice that it is also a removable discontinuity because $x+1$ is the common factor in the numerator and denominator.
This graph also has a removable discontinuity at $x=-1$ because $x+1$ is the common factor in the numerator and denominator of the rational function.
An asymptote of a graph is an imaginary line that the graph gets close to as $x$ goes to plus or minus infinity or a particular number. For example, the graph of the rational function $f(x)=x1 $ has two asymptotes — the $x-$axis and the $y-$axis.
Analyzing the diagram, the following can be observed.
To identify the vertical asymptotes, the function should be in its simplest form. That is, if $p(x)$ and $q(x)$ have common factors, the function should be simplified first. The vertical asymptotes will occur at the zeros of the denominator. Check out the following examples.
The degrees of polynomials in the numerator and denominator of a rational function will determine if the graph has a horizontal asymptote.
$f(x)=q(x)p(x) =bx_{n}+…ax_{m}+… $ | |
---|---|
Horizontal Asymptote | If $m>n,$ there is no horizontal asymptote. |
If $m<n,$ the horizontal asymptote is $y=0.$ | |
If $m=n,$ the horizontal asymptote is $y=ba .$ |
Vertical Asymptotes: $t=-2$ and $t=-1$
Horizontal Asymptote: $y=0$
Factor the numerator. What does the Factor Theorem state?
Factor Theorem |
The expression $x−a$ is a factor of a polynomial if and only if the value $a$ is a zero of the related polynomial function. |
$t$ | $t_{3}−t_{2}−10t−8$ | $p(t)=t_{3}−t_{2}−10t−8$ |
---|---|---|
$1$ | $1_{3}−(1)_{2}−10(1)−8$ | $-18$ |
$-1$ | $-1_{3}−(-1)_{2}−10(-1)−8$ | $0$ |
$2$ | $2_{3}−(2)_{2}−10(2)−8$ | $-24$ |
$-2$ | $-2_{3}−(-2)_{2}−10(-2)−8$ | $0$ |
$4$ | $4_{3}−6(4)_{2}−7(4)+60$ | $0$ |
$t_{3}−t_{2}−10t−8=(t+1)(t+2)(t−4)$
Cross out common factors
Cancel out common factors
$y=bx_{n}ax_{m} $ | Asymptote |
---|---|
$m<n$ | $y=0$ |
$m>n$ | none |
$m=n$ | $y=ba $ |
An asymptote can be neither vertical nor horizontal. It can be a slanted line. In such cases, it is called an oblique asymptote or slant asymptote.
An oblique asymptote of a rational function occurs when the degree of the numerator is $1$ more than that of the denominator. Therefore, a function with an oblique asymptote can never have a horizontal asymptote. To find the equation of the oblique asymptote of a rational function $f(x)=q(x)p(x) ,$ $p(x)$ is divided by $q(x)$ using long division to get a quotient $ax+b$ with a remainder $r(x).$$4x_{2}x_{3} =4x $
Multiply $term$ by $divisor$
Subtract down
The students spent about $3$ hours at the factory. At the end of this period, LaShay's teacher made a survey about the trip and drew a graph relating the students' interest in the trip and the time elapsed at the factory. The greater the $S(t)$ value, the greater the interest of the students.
The graph above is the part of the graph of the following function in the first quadrant.Vertical Asymptotes: $t=-1$
Oblique Asymptote: $O(t)=-2t +2$
Start by subtracting the rational expressions on the right-hand side of the function.
$ba =b⋅2(t+1)a⋅2(t+1) $
Multiply
Subtract fractions
Distribute $2t$
Commutative Property of Addition
$(a+b)_{2}=a_{2}+2ab+b_{2}$
Distribute $2$
$2t_{2}-t_{3} =-2t $
Multiply $term$ by $divisor$
Subtract down
$2t_{2}4$