I have f(x) + 1/ x-c this is the question and I am told for the function with a vertical asymptote at x=1
Mr. Reimy was kind enough to provide this information
comparing it to y = 1/x
for the first interpretation the graph of y = 1/x would simply be translated vertically by c
for the 2nd interpretation the graph of y = 1/x would be translated horizontally by -c , e.g
if y = 1/(x+2) is a shift of 2 to the left
if y = 1/(x-5) is a shift of 5 to the right
I now see that the c is responsible for the vertical shift so substituting 1 in for x gives me a linear graph that goes through the point (1,0) correct?
if f(x) = 1/(x-c), then there is a vertical asymptote at x=c. This is the graph of f(x) = 1/x, shifted right by c.
If f(x) = 1/x - c then the graph is f(x)=1/x, shifted down by c, and the vertical asymptote is at x=0.
And you have your shifts backwards.
Plugging in x=1 in either case does not give you (1,0). You need some serious review of your algebra.