At what value of x does the graph of the following function f(x) have a vertical asymptote? f(x)=1/x-1

Vertical asymptotes are the x-values for which a particular function is undefined;
With linear reciprocal functions like the one given, it will be when the denominator equates to 0 as it is impossible to divide any number by 0;
So we just have to equate the denominator to 0 and rearrange to give x:
x - 1 = 0
x = 1
The line x = 1 is the vertical asymptote for this function.

Horizontal asymptotes are the y-values for which a particular function is undefined;
Finding the horizontal asymptote for linear reciprocal functions is quite simple;
For this kind of function, where there is no x term in the numerator, the horizontal asymptote is just equivalent to the constant added to the fraction (note: having no added constant is the same as having an added 0);
So:
F(x) = 1/(x - 1) = (1/(x - 1)) + 0
The horizontal asymptote is y = 0 for this function.

If it was F(x) = (1/(x - 1)) + 1, the horizontal asymptote would be y = 1.

Idon’t exactly understand but i think dbc=rst