Find the vertical asymptotes, if any, of the graph of the rational function. Show your work.
f(x) = (x-4)/(x(x-4))
(x-4)/x(x-4) the common factors cancel out and all is left is f(x)= 1/x...
how do I solve this problem?
vertical asymptotes occur where the denominator is zero and the numerator is not.
So, your graph has a vertical asymptote at x=0
try entering your function at wolframalpha.com to see its graph, and other information.
Note that as long as x≠4, f(x) = 1/x
There is a hole at x=4, because then f(x) = 0/0 which is undefined.
To find the vertical asymptotes of the rational function f(x) = (x-4)/(x(x-4)), we need to determine the values of x for which the function becomes undefined. In other words, we need to find the values of x that would make the denominator (x(x-4)) equal to zero.
In this case, we have two factors in the denominator: x and (x-4). Setting each factor equal to zero, we get the following equations:
1) x = 0
2) x-4 = 0
Solving these equations gives us the values of x for which the function becomes undefined.
1) For x = 0, the denominator becomes zero, meaning the function blows up at x = 0. Thus, x = 0 is a vertical asymptote.
2) For x = 4, the denominator also becomes zero, indicating another vertical asymptote at x = 4.
So, the vertical asymptotes of the graph of f(x) = (x-4)/(x(x-4)) are x = 0 and x = 4.
Note: It's important to consider the fact that the common factors canceled out and simplified the function to f(x) = 1/x. However, we need to be cautious when canceling common factors, as we might miss out on potential asymptotes that would appear in the original form of the function.