First, factorize the function.
f(x) = (3x-2)(x+1)
(2x+5)(x+1)
Simplify to a linear relationship
f(x) = (3x-2)
(2x+5)
x cannot be -1 and -5/2.
Noticed that the graph has only one vertical asymptote, that is x=-5/2.
This is a special case because of a line that is discontinuous at x=-1.
This is the value at which the function is undefined.
Here, the discontinuity is not an asymptote but the point (-1, -5/3).
This type of discontinuity is called a hole or a gap.
To solve rational inequalities, it helps by rewriting with the right side equal to zero. A number line can be used to consider intervals. Then pick a number of each interval to determine the sign of the factors.
For example, x/x+1 < 2x/x-2
x/x+1 - 2x/x-2 <0
x(x-2) - 2x(x+1) < 0 (x-2)(x+1)
x^2-2x-2x^2-2x < 0 (x-2)(x+1)
-x^2 - 4x < 0
(x-2)(x+1)
-x ( x+4 ) < 0
(x-2)(x+1)
Restriction of x: x cannot be -1 and 2.
1. List all the factors.
2. Use number line/table to help to consider each intervals.
3. Pick a number and test the value.
Factors / Intervals
x < -4
-4 < x < -1
-1 < x < 0
0 < x < 2
x > 2
Test Values
-5
-2
-0.5
1
3
-x
+
+
+
-
-
x+4
-
+
+
+
+
x-2
-
-
-
-
+
x+1
-
-
+
+
+
-
+
-
+
-
x < -4, -1 < x < 0 and x > 2 are negative or less than zero.
Hence the solution of x/x+1 - 2x/x-2 <0 arex < -4, -1 < x < 0 and x > 2.
For this part, we will focus more on how to find the horizontal asymptote of the function.
The equation of the horizontal asymptote can be found by dividing each term in both the numerator and denominator by x and investigating the behaviour of the function as x approaches to positive or negative infinity.
Short cut to find the horizontal asymptote of the function:
The equation of the horizontal asymptote is same as the coefficient x in the numerator divided by the coefficient s in the denominator.
Thus, horizontal x is a/c.
Isn't it way way easier than dividing each term by x?
NOTES:
The coefficient b acts to stretch the curve but no effect on the asymptotes, domain or range.
The coefficient d shifts the vertical asymptote.
The two brances of the graph of the function are equidistant from the point of intersection of the vertical and horizontal asymptotes.
In this section, we will analyse reciprocal of quadratic function such as f(x)= 5/ x^2 - 4x + 3.
Reciprocal of quadratic functions with two zeroes have three parts, with the middle one reaching a maximum or minimum point. This point is equidistant from the two vertical asymptotes.
The behavior of the function near the asymptotes is similar to the previous post, that is reciprocal of linear functions.
We can predict the key features of the graph by analyzing the roots of the quadratic relation in the denominator.
For example:
The denominator of the above function is x^2 - 4x + 3.
First, we have to factorize the relation. It is (x - 3)(x - 1).
The denominator cannot equal to zero. Thus the restriction of x are 3 and 1. They will be the vertical asymptotoes of the function.
To find the key features such as domain, range, end behavior and asymptotes, please refer to the previous post.
There are 5 intervals in the graph.
For the interval x<1, the sign of f(x) is positive. The sign of slope is positive and the change in slope is increasing.
For the interval 1<x<2, the sign of f(x) is negative. The sign of slope is positive and the change of slope is decreasing.
For the interval x=2, the sign of f(x) is 0. The sign of slope is positive and the change of slope is constant.
For the interval 2<x<3, the sign of f(x) is negative. The sign of slope is negative and the change of slope is decreasing.
For the interval x>3, the sign of f(x) is positive. The sign of slope is negative and the change of slope is increasing.
Source:
McGraw-Hill Ryerson Advanced Functions 12
For further details and understanding, you can watch these videos :)
The reciprocal of a linear function has the form f(x) = 1/kx-c.
Let use this function as example for better understanding:
f(x) = 1/ x-4
Domain
To find the domain of the function, we have to know the restriction of the x first.
It's easy to find the restriction of x. All you have to do is let the denominator equal to zero.
x-4=0
x=4
So x cannot equal to 4. If it is equal to 4, the function will be undefined.
So the domain is { x € R; x ≠ 4}.
End Behaviour
As x approaches 4 from the right, f(x) approaches positive infinity.
As x approaches 4 from the left, f(x) approaches negative infinity.
The graph approaches the vertical line x=4 but does not cross it. So the curve is discontinuous at this line. This line is called vertical asymptote.
As x approaches positive infinity, f(x) approaches zero from the above as all the values of f(x) are positive.
As x approaches negative infinity, f(x) approaches zero from below as all the values of f(x) are negative.
The graph approaches a horizontal line at x-axis but does not cross it. It has the horizontal asymptote of y=0.
Vertical Asymptote
The graph will not cross the line x=4.
Thus, an equation for the vertical asymptote is x=4.
Horizontal Asymptote
The graph will not cross the line y=0.
Thus, an equation for the horizontal asymptote is y=0.
Range
Same as domain, we have to find the restriction of y first.
The graph of the function shows that f(x) gets close to the line y=0 but never actually touches the line. Therefore the only restriction on the range of f(x) is y not equal to 0.
