Content Curator
Rational Functions are those functions that consist of the algebraic ratio of two polynomial functions where the denominator is not equal to zero. The domain of a rational function is the set of all natural numbers except the value of zero in the denominator. A function of one variable, a, is called a rational function if, it can be represented as f(a) = p(a)/q(a), where p(a) and q(a) are polynomials such that q(a) ≠ 0.
Also check: NCERT Solutions for Relations and Functions
| Table of Content |
Key Terms: Rational Function, Asymptotes, Polynomial, Denominator, x-intercept
What are Rational Functions?
[Click Here for Sample Questions]
When a polynomial function is divided by another polynomial function, the result is known as a rational function provided that the denominator of such a ratio is not equal to zero.
R(x) = \(\frac{P(x)}{Q(x)}\), where Q(x) is not equal to 0.
If p = P(x) and q = Q(x)
and
R(x) is equal to \(\frac{P(x)}{Q(x)}\)
where q is not equal to 0,
then, R(x) is a Rational Runction.
Examples of Rational Functions
A few examples of rational functions are provided below:
| \(y = \frac{1}{x}, y= \frac{x}{x^2-1}, y=\frac{3}{x^4+2x+5}\) |
Also check:
Asymptotes of Rational Functions
[Click Here for Sample Questions]
There are three types of asymptotes present in the rational function.
- Vertical Asymptotes
- Horizontal Asymptotes
- Oblique Asymptotes
Vertical Asymptote
R(x) will have the vertical asymptote when the value of Q(x) equals to zero.
Horizontal Asymptote
When the degree of the numerator P(x) is less than the degree of the denominator Q(x), then the rational function will have a horizontal asymptote.
Oblique Asymptote
When the degree of the numerator Q(x) is so greater than zero that the value of R(x) becomes somewhat equal to the curve T(x) of the function, then, the rational function will have an oblique asymptote.
How to Graph a Rational Function?
[Click Here for Sample Questions]
The steps to graph a rational function are mentioned below:
- Find out the Asymptote of the Rational Function.
- Draw the Asymptote as dotted lines.
- Find out the x-intercept and y-intercept of the function.
- Find the values of y with the different values of x.
- Plot the points on a graph and draw a smooth line to join the points. Make sure that the lines do not cross the asymptotes.
Solved Examples
[Click Here for Sample Questions]
Example 1. Draw a suitable graph for the rational function below.
\(y = \frac{4x+1}{2x+1}\)
Solution. The vertical asymptote is represented by the value of x.
When the denominator is zero, the value of x can be calculated by equating the polynomial function in the denominator by considering its value to be zero.
2x+1 = 0
x = \(-\frac{1}{2}\)
Therefore, the value of x = -0.5
This gives the value of the vertical asymptote to be -0.5
Now, the asymptote can be expressed in the form of a dotted line on the graph as shown below:
By equating the function by different values of x, the x-intercept and the y-intercept can be determined and various points can be plotted on the graph.
The x-intercept is at (-0.25, 0) and the y-intercept is at (0, 1).
The following graph is plotted by joining the various points on the graph:
Sometimes the given rational function has to be simplified before graphing it. In that case, there will be additional steps for the additional values.
Example 2. Find the vertical asymptote of the function f(x)= (x²+5x+6)/(x²+x-2).
Solution. The given function, on simplification, gives the equation f(x)= (x+3)/(x-1).
On considering the denominator to 0, we get
x-1 = 0
x = 1
Thus, the vertical asymptote of the given equation is x = 1.
Also check:
Things to Remember
- The denominator of the algebraic ratio in a function that is rational can never be zero.
- The asymptotes of rational functions are- horizontal asymptotes, vertical asymptotes and oblique or slant asymptotes.
- The vertical asymptote can be calculated by equating the polynomial denominator with a null value.
- Sometimes the given rational function must be equated to make it simple, thereby, giving rise to additional steps.
Sample Questions
Ques 1. What will be the horizontal asymptote of the function f(x) = (x²+5x+6)/(x²+x-2)? (2 Marks)
Ans. The degree of the numerator N = 2 and the degree of the denominator D = 2
The horizontal asymptote is given by y = 1/1 = 1
Therefore, the horizontal asymptote is y = 1
Ques 2. What is the slant or oblique asymptote of the given function f(x)= x²/(x+1)? (2 Marks)
Ans. The degree of numerator N is 2.
The value of the denominator D is 1.
On dividing x² by (x+1) , we get the quotient as (x-1).
Therefore, the slant asymptote is y= x-1
strong>Ques 3. Find the horizontal and vertical asymptotes of the rational function: f(x) = (3x3 - 6x) / (x2 - 5). (3 Marks)
Ans. The function is in the simplest form.
For finding the vertical asymptote, let x² - 5 = 0. Solving this, we get x = ± √5.
Since the degree of the numerator (3) > degree of the denominator (2), it has no horizontal asymptote.
Therefore, the vertical asymptotes are at x = √5 and x = -√5 and there is no horizontal asymptote.
Ques 5. Sketch the graph of the following function. Clearly identify all intercepts and asymptotes. (5 Marks)
\(f(x)= \frac{-4}{x-2}\)
Ans. The y-intercept is the point (0,f(0))= (0, 2). The numerator is a constant. So, it can never be zero. We assume the denominator to be zero. So we get the following
x – 2 = 0 \(\rightarrow\) x =2
So, we’ll have a vertical asymptote at x= 2
For this equation the largest exponent of x in the numerator is zero since the numerator is a constant. The largest exponent of x in the denominator is 1, which is larger than the largest exponent in the numerator, and so the x-axis will be the horizontal asymptote. we only have one vertical asymptote and so we only have two regions to our graph: x<2 and x>2. We’ll need a point in each region to determine if it will be above or below the horizontal asymptote. Here are a couple of function evaluations for the points.
\(f (0) = 2 \rightarrow (0,2)\)
\(f (3) = -4 \rightarrow (3,-4)\)
Here is a sketch of the function with the points found above. The vertical asymptote is indicated with a blue dashed line and recall that the horizontal asymptote is just the x-axis.
Ques 6. Sketch the graph of the following function. Clearly identify all intercepts and asymptotes. (5 Marks)
\(f(x) = \frac{6-2x}{1-x}\)
Ans. The y-intercept is the point (0,f(0))= (0, 6). For the x-intercepts we set the numerator equal to zero and solve. Doing that for this problem gives,
\(f(x) = \frac{6-2x}{1-x}\)
So, the only x-intercept for this problem is (3,0). We can find any vertical asymptotes be setting the denominator equal to zero and solving.
1 – x = 0 \(\rightarrow\) x = 1
So, we’ll have a vertical asymptote at x= 1. For this equation the largest exponent of x in both the numerator and denominator is 1. Therefore, the horizontal asymptote for this problem is then the coefficient of the x in the numerator divided by the coefficient of the x in the denominator. Or,
\(y = \frac{-2}{-1} = 2\)
we only have one vertical asymptote and so we only have two regions to our graph : x<1 and x>1. We’ll need a point in each region to determine if it will be above or below the horizontal asymptote. Here are a couple of function evaluations for the points.
\(f (0) = 6 \rightarrow (0,6)\)
\(f (3) = 0 \rightarrow (3,0)\)
Here is a sketch of the function with the points found above. The vertical and horizontal asymptotes are indicated with blue dashed lines.
Ques 7. Sketch the graph of the following function. Clearly identify all intercepts and asymptotes. (5 Marks)
\(f (x) = \frac{8}{x^2+x-6}\)
Ans. The y-intercept at the point(0,f(0))= (0,-4/3). For the x-intercepts, we set the numerator equal to zero and solve. However, in this case, the numerator is a constant (8 specifically) and so can’t ever be zero. Therefore, this function will have no x-intercepts. We can find any vertical asymptotes by setting the denominator equal to zero and solving. Doing that for this function gives,
So, we’ll have two vertical asymptotes at x= -3 and x= 2. For this equation, the largest exponent of x in the numerator is zero since the numerator is a constant. The largest exponent of x in the denominator is 2, which is larger than the largest exponent in the numerator, and so the x-axis will be the horizontal asymptote. we only have two vertical asymptotes and so we have three regions to our graph :
x < –3, –3 < x, < 2 and x > 2
We’ll need a point in each region to determine if it will be above or below the horizontal asymptote. There are a couple of possible different behaviors in the middle region. To determine just what the behavior is we need to get a couple of points in this region. The best idea for points in the middle region is to check a couple of points close to the vertical asymptotes we know if the edge is going to be above or below the horizontal asymptote. Here are some function evaluations for the points.
From the second and third points we see that the curve in the middle region should be below the horizontal asymptote (x-axis for this problem) at both edges and so the curve will be completely below the horizontal asymptote in this whole region.Here is a sketch of the function with the points found above. The vertical asymptote is indicated with a blue dashed line and recall that the horizontal asymptote is just the x-axis.
Check out:
Comments