Given the function: f(x) = 5/(x-1)^2
Find the
Horizontal Asymptote:
Vertical Asymptote:
Hole(s):
X intercept:
Y intercept:
Here is a "picture" of your graph
It should help you to answer your questions.
http://www.wolframalpha.com/input/?i=plot+y%3D+5%2F%28x-1%29%5E2+
To find the horizontal asymptote, we need to determine the behavior of the function as x approaches positive or negative infinity.
The given function is f(x) = 5/(x-1)^2. As x approaches positive or negative infinity, the denominator (x-1)^2 becomes very large, resulting in a very small value for the function. Therefore, the numerator 5 becomes insignificant compared to the denominator, and we can say that the function approaches zero as x approaches positive or negative infinity.
Hence, the horizontal asymptote is y = 0.
To find the vertical asymptote, we need to examine the behavior of the function as x approaches the values that make the denominator equal to zero.
In this case, the denominator (x-1)^2 becomes zero when x = 1. Therefore, there is a vertical asymptote at x = 1.
To find the holes in the graph, we need to identify the values of x that make both the numerator and denominator of the function equal to zero.
In this case, the numerator 5 can never be equal to zero. However, the denominator (x-1)^2 becomes zero when x = 1. Therefore, the function has a hole at x = 1.
To find the x-intercept, we need to determine the values of x for which the function equals zero.
In this case, to find the x-intercept, we set the function f(x) = 5/(x-1)^2 equal to zero and solve for x:
5/(x-1)^2 = 0
Since we cannot have a fraction equal to zero, there is no x-intercept for this function.
Lastly, to find the y-intercept, we can substitute x = 0 into the function and calculate the corresponding y-value.
f(0) = 5/(0-1)^2
f(0) = 5/1 = 5
Therefore, the y-intercept is (0,5).