Question

Which of the following is true about the graph of f(x)=x-2/x^2-2x


Vertical Asymptote at x = 0; Slant Asymptote at y = x

Vertical Asymptote at x = 0; Horizontal Asymptote at y = 1

Vertical Asymptote at x = 0; Hole at x = 2; Horizontal Asymptote at y = 1

Vertical Asymptote at x = 0; Hole at x = 2; Horizontal Asymptote at y = 0

Answer 1

well if you factorise the denominator you get

\[f(x) = \frac{x -2}{x(x - 2)}\]

which can be simplifed to f(x) = 1/x

so the curve has a vertical asymptote at x = 0 and horizontal asymptote at y = 0


by x cannot equal 2 since it will result in a zero denominator... in the original function

but since (x -2) can be cancels then x = 2 is a point of discontinuity. (a Hole)

Answer 2

so...whats the answer?

Answer 3

did you read the information..? as the answer is in there...