William W. answered • 06/21/20
Experienced Tutor and Retired Engineer
I'm assuming that the function in question is:
We find vertical asymptotes by looking at the denominator (4x + 1). Since division by zero is undefined, the denominator cannot equal zero. So to find the vertical asymptote (the values that "x" cannot be) we can set the denominator equal to zero to see what the issues are. (please note that there are cases when the denominator = 0 that result in a hole in the graph instead of a vertical asymptote. These occur when there is a factor in the denominator that cancels out a factor in the numerator. This does not occur in this example.) So:
4x + 1 = 0
4x = -1
x = -1/4
Meaning x cannot be -1/4 or, in other words, there is a vertical asymptote at x = -1/4
To find the horizontal asymptote, we consider what f(x) approaches as x gets very large. For instance:
When x = 100, f(100) = (100 - 1)/(4•100 + 1) = 99/401 = 0.24688
When x = 1000, f(1000) = (1000 - 1)/(4•1000 + 1) = 999/4001 = 0.24969
When x = 10000, f(10000) = (10000 - 1)/(4•10000 + 1) = 9999/40001 = 0.24997
When x = 100000, f(100000) = (100000 - 1)/(4•100000 + 1) = 99999/400001 = 0.249997
Obviously, f(x) approaches 0.25 as x approaches ∞ meaning the horizontal asymptote is y = 0.25
Another way of doing this for polynomials with the same degree in the numerator and denominator is to divide the two leading coefficients so 1/4.
There is no slant asymptote. Slant asymptotes occur when the numerator has a greater degree than the denominator.
The x-intercepts occur when f(x) = 0, when we set f(x) = 0, the only part of the function that can make that happen is the numerator, so we just set x - 1 equal to zero:
x - 1 = 0
x = 1 is the x-intercept (so (1, 0)
The y-intercept occurs when x = 0 so we just plug in zero for x:
f(0) = (0 - 1)/(4•0 + 1)
f(0) = -1/1
f(0) = -1
So the y-intercept is 0 (so 0, -1)
