Question

please check my work..!

Answer 1

I have no idea what I'm doing

Answer 2

The removable discontinuity of the graph is the point that is not included... In this instance what can x not equal?
For the x intercepts- let y = 0 and solve. For the y intercepts let x= 0 and solve

Answer 3

To solve for the removable discontinuity of this particular graph - you must find the limitations of the domain. We know there is a limitation on domain as we are working with a fraction. We cannot divide by zero... So you need to find the number that would make the denominator zero. This is the removable discontinuity

Answer 4

Now- looking at the y-intercepts.... If we were to let x = 0 and solve we'd find that 0 is our removable discontinuity as well as our vertical asymptote... So there are no y-intercepts

Answer 5

If we were to let y = 0 and solve for our x intercepts we could get -2 and 10

Answer 6

@sunnnystrong I thought the removable discontinuity only occurs when the numerator and the denominators are equivalent..?

Answer 7

also, shouldn't this be a slant asymptote?

Answer 8

I think the factoring is incorrect, but I don't know how critical that is to the rest of the question.

(x-10)(x-2) = x^2 - 12x + 20