Sketching Of Rational Functions
What is a rational function?
A rational function is one which has a polynomial in the numerator and another polynomial in the denominator. For example, (x^2)/(x^3 + 1) (read 'x' squared and 'x' cubed), is a rational function whereas (x^0.5)/(x^3 + 1) (x^0.5 is square root of 'x') is not, since the numerator is not a polynomial.
Now, let us make a list of the information that we have to collect in order to sketch a graph of a rational function.
Find the x-intercept/s or zero/s.
Find the y-intercept/s.
Find the Vertical Asymptote/s.
Find the Horizontal Asymptote.
Find the Oblique Asymptote (if any).
Draw the sign diagram of the function to understand where the function is above and where it is below the x-axis.
Find the maxima, the minima and the stationary inflection points.
Find the intervals where the function is increasing.
Find the intervals where the function is decreasing.
Find the intervals where the function is concave upwards.
Find the intervals where the function is concave downwards.
Find the non-stationary points of inflection.
Sketch the graph.
The steps that I have listed above are for those rational functions which do not have any common factor/s between the numerator and the denominator. In case, they have any common factor/s, there will be holes present in their graphs and after cancellation of the common factor/s there is no guarantee that the final expression for the function will be a rational function.
For example, f(x) = (x^2 - 1)/(x + 1)
We find both the numerator and the denominator have the common factor (x + 1) since the numerator is equal to (x + 1)(x - 1). Therefore, after cancellation of the common factor, it becomes a linear function and its graph is a straight line with the only exception that there is a hole in the line at the point (-1 , -2). The x-coordinate of the hole is obtained by equating the common factor to 0. The y-coordinate of the hole is obtained by substituting the x-coordinate value in the function expression (after cancellation of the common factor). If there are multiple common factors, there will be multiple holes.
Let us now come back to our discussion on sketching of rational functions.
Example 1 f(x)= (2x + 6)/(x + 2)
1. Finding the x-intercept
We find the x-intercept/s by equating the function to zero.
Therefore, (2x + 6)/(x + 2) = 0 which means 2x + 6 = 0 or x = -3.
So, -3 is the zero of this function or, in other words, this function passes through the point (-3 , 0).
2. Finding the y-intercept
We find the y-intercept/s by getting the value of the function when x = 0, which means we substitute the value of 0 in place of 'x' and find the corresponding value of the function.
Therefore, y = (2 . 0 + 6)/(0 + 2) = 3.
So, the y-intercept of the function is 3 or the function passes through the point (0 , 3).
3. Finding the Vertical Asymptote
We find the vertical asymptote/s of a rational function by getting the value/s of 'x' for which the denominator becomes equal to zero.
The significance of the vertical asymptote is that the function will never meet the vertical asymptote line and as 'x' approaches the vertical asymptote value either from the left or from the right, the function approaches positive infinity or negative infinity.
Therefore, we equate the denominator of this function to 0.
x + 2 = 0 which means x = -2 is the vertical asymptote of this function.
4. Finding the Horizontal Asymptote
We may find the horizontal asymptote of any function in several ways but the most commonly used method is --
a) When the highest power of 'x' in the numerator is equal to the highest power of 'x' in the denominator
The equation of the horizontal asymptote is y = (coefficient of the highest power of 'x' in the numerator)/(coefficient of the highest power of 'x' in the denominator).
b) When the highest power of 'x' in the numerator is less than the highest power of 'x' in the denominator
The equation of the horizontal asymptote is y = 0.
The significance of the horizontal asymptote is that the function will never meet the horizontal asymptote line and as 'x' approaches positive or negative infinity, the function approaches the horizontal asymptote line from below or above.
Our function satisfies condition a), since both the numerator and the denominator are linear polynomials, hence the highest power of 'x' is the same, 1.
Therefore, the horizontal asymptote equation is y = 2/1 (2 divided by 1) or y = 2.
5. Finding the Oblique Asymptote
We have an oblique asymptote for a rational function when the highest power of 'x' in the denominator is one less than the highest power of 'x' in the numerator.
Since we do not have such a situation in our function, therefore, our function does not have any oblique asymptote.
6. Drawing the sign diagram of the function
Thus, we see that our function is above the x-axis when x < -3 and when x > -2 and it is below the x-axis when -3 < x < -2. It is equal to 0 when x = -3.
7. Finding the maxima, the minima and the stationary inflection points
We find the maxima and the minima points of the function by first of all, getting its first derivative and then equating it to 0. The 'x' values thus obtained correspond to either the 'x' coordinate of the maximum point, the minimum point or the stationary inflection point on the graph.
We find f'(x) is never equal to 0. Therefore, there is no maxima, minima or stationary point of inflection for this function.
8. Finding the intervals where the function is increasing
The function is increasing in the region where f'(x) is positive.
For this, we need to draw the sign diagram of the f'(x) function.
From the sign diagram we find the function is not increasing in the entire domain.
9. Finding the intervals where the function is decreasing
From the sign diagram of f'(x), we find that the function is decreasing on both sides of the vertical asymptote x = -2.
10. Finding the intervals where the function is concave up
To find the shape of the curve, we have to first find the second derivative of the function and then draw its sign diagram. The regions where the second derivative, f''(x) is positive, the function is concave up.
We find that f''(x) is never equal to 0.
From, the sign diagram of f''(x), it is clear that the function f(x) is concave up for x > -2.
11. Finding the intervals where the function is concave down
The function is concave down where f''(x) < 0, therefore for our function, it is concave down when x < -2.
12. Finding the non-stationary points of infection
Non-stationary points of inflection are those points on the graph whose x-coordinates make f'(x) != 0 (read 'not equal to'), f''(x) = 0 and around which the sign of f''(x) changes.
Since f''(x) != 0 for any value of 'x', therefore we do not have any non-stationary point of inflection.
13. Sketching of the graph
Example 2 f(x)= (3x - 9)/(x^2 - x - 2)
1. Finding the x-intercept
(3x - 9)/(x^2 - x - 2) = 0 which means 3x - 9 = 0 (the numerator should be equal to 0 in order for the function to have a value of 0). So, the x-intercept or the zero of the function is 3. The function passes through the point (3 , 0).
2. Finding the y-intercept
The y-intercept of this function is (3 . 0 - 9)/(0^2 - 0 - 2) or 4.5. The function passes through the point (0 , 4.5).
3. Finding the Vertical Asymptotes
x^2 - x - 2 = 0. On factorizing the expression, we have (x + 1)(x - 2) = 0. We find the L.H.S. becomes equal to zero when x = -1 or x = 2. Therefore, x = -1 and x = 2 are the two vertical asymptotes of this function.
4. Finding the Horizontal Asymptote
For our function we find, the numerator to be a linear polynomial, hence the highest power of 'x' is 1, and the denominator to be a quadratic polynomial, hence the highest power of 'x' is 2. Therefore, the highest power of 'x' in the numerator is less than that in the denominator, hence the horizontal asymptote for our function is y = 0.
5. Finding the Oblique Asymptote
Since we do not have the highest power of 'x' in the numerator greater than the highest power of 'x' in the denominator of our function, our function does not have any oblique asymptote.
6. Drawing the sign diagram of the function
Therefore, the function is below the x-axis for x< -1, is above the x-axis in the region -1 < x < 2, again below the x-axis between 2 < x < 3, equal to 0 at x = 3 and finally above the x-axis when x > 3.
7. Finding the maxima, the minima and the stationary inflection points
In order to find out which x-coordinate is corresponding to which type of point, we draw the sign diagram of f'(x).
If the sign of f'(x) changes from negative to positive around a point where f'(x) = 0, then that point corresponds to a minima.
If the sign of f'(x) changes from positive to negative around a point where f'(x) = 0, then that point corresponds to a maxima.
If the sign of f'(x) does not change (or remains the same) around a point where f'(x) = 0, then that point corresponds to a stationary inflection point.
We may classify the points using the second derivative expression for the function as well.
If, for a particular 'x' value for which f'(x) = 0, the second derivative, f''(x) < 0, then that point is a maxima but if f''(x) > 0, then that point is a minima. But if both f'(x) = 0 and f''(x) = 0 then it is a stationary point of inflection.
We find that the sign of f'(x) changes from negative to positive around the point x = 1, therefore, this point corresponds to the minima of the function.
The sign of f'(x) changes from positive to negative around x = 5, therefore this point corresponds to the maxima of the function.
The y-coordinates of the maxima and the minima points are found by substituting x = 5 and x = 1 in the function expression and getting the corresponding values of the function.
Therefore, the function has a minima (or local minima) at (1 , 3) and a maxima (local maxima) at (5 , 1/3).
This function does not have any stationary inflection point.
8. Finding the intervals where the function is increasing
Our function is increasing (found from the sign diagram of f'(x)) between 1 <= x < 2 ( read as 'x' greater than or equal to 1) and between 2 < x <= 5
9. Finding the intervals where the function is decreasing
The intervals in which the function is decreasing is found from the sign diagram of f'(x). The regions where the value of f'(x) is negative are the intervals where the function is decreasing. Therefore, our function is decreasing when x < -1, between -1 < x <= 1 and x>= 5.
10. Finding the intervals where the function is concave up
It is difficult to find the zeros of f''(x) without a GDC but we can take the asymptotes and try to draw the sign diagram of f''(x).
Therefore, the function is concave up in the interval -1 < x < 2.
11. Finding the intervals where the function is concave down
From the sign diagram of f''(x), we find that the function is concave when x < -1 and when x > 2.
12. Finding the non-stationary points of infection
Non-stationary points of inflection are those points on the graph whose x-coordinates make f'(x) != 0 (read 'not equal to'), f''(x) = 0 and around which f''(x) changes sign.
As I have mentioned before, finding the zero of f''(x) for this function is extremely difficult without a calculator and we will see we do not need the knowledge of this for sketching the graph.
13. Sketching the graph
Example 3 f(x)= - x/(x^2 - 4)
1. Zero of the function x = 0
2. y-intercept y = 0. That means, the function passes through the origin (0 , 0).
3. Vertical Asymptotes x = -2 and x = 2
4. Horizontal Asymptote y = 0
5.
The sign diagram of f(x) shows us that the function is above the x-axis for intervals -2 < x < 0 and x > 2 and is below the x-axis in the intervals x < -2 and 0 < x < 2.
6.
We find, f'(x) does not have any real zeros. Therefore, the function does not have any maxima or minima points.
7.
From the sign diagram of f'(x), we find that the function is decreasing throughout the entire domain.
8.
f''(x) = 0 when x = 0, 2 or -2. But x = 2 and x = -2 are the two asymptotes.
9.
Since the sign of f''(x) is changing around the point x = 0, therefore, it is the non-stationary point of inflection. The curve is concave up in the intervals -2 < x < 0 and x > 2 and is concave down in the intervals x < -2 and 0 < x < 2.
10. The graph
Example 4 f(x)= x + 1/x = (x^2 + 1)/x
1. There is no zero of the function, since x^2 + 1 = 0 does not give us any real value of 'x' as the solution.
2. There is no y-intercept of the function.
3. x = 0 is the vertical asymptote.
4. We don't have any horizontal asymptote for this function. But we have an oblique asymptote because the highest power of 'x' in the numerator is greater than that of the denominator. For finding the equation of the horizontal asymptote, we divide numerator by the denominator and the quotient of the division gives us the expression for the oblique asymptote.
Therefore, y = x is the oblique asymptote.
5.
The function is below the x-axis on the left side of the y-axis and is above the x-axis on the right hand side of the y-axis.
6.
Therefore, the function is decreasing in the intervals -1 <= x < 0 and 0 < x <= 1 and increasing in the intervals x <= -1 and x >= 1. We have the maxima of the function at x = -1 and the minima of the function occurs at x = 1. (-1 , -2) and (1 , 2) are the coordinates of the maxima and the minima of the function respectively.
7.
We find that the function is concave down to the left hand side of the origin and is concave towards the right hand side of the origin.
8.
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