For the first one, something like
y = (x-2)/(x^2 - 4) should do it
for the second, how about
y = x^2(x-4)/((x-2)(x+3)(x-4) )
what is a function that Contains no vertical asymptotes but has a hole at x=2 and
another function that contains a horizontal asymptote of 1, vertical asymptotes of 2 and -3, and a hole at x=4.
3 answers
thank you
no vertical asymptotes but has a hole at x=2
y = (x^2-2x)/(x-2)
y = (x^2-2x)/(x-2)