Here's the question you clicked on:
raytiller1
Consider the function f(x) = x2-x-12/x-4 . Describe the graph of this function. Include all discontinuities, intercepts, and the basic shape of the graph.
Consider the expression: (x^2-x-12)/(x-4)=(x-4)(x+3)/(x-4) =(x-3) after cancelling the common factor for points where x not equal to 3. At x=3, the function is undefined. Is that enough to get you gonig?
You can't cancel and forget about the x-4 term. There's still a point discontinuity there.
But otherwise it behaves like the line.
now im really confused @mathmate @vf321
Sorry, I meant to say that the function is undefined at x=4 (where we cancelled). The remaining points behave like y=x+3.
@raytiller1 What mathmate's saying is right - If you notice, \[f(x)=\frac{x^2-x-12}{x-4} = \frac{(x-4)(x+3)}{x-4}\] So, when x!=4, you have some number x-4 on the numerator and in the denominator, so they cancel. \[x \neq 4::f(x) = \frac{x-4}{x-4}(x+3)=1(x+3)\] When x does equal 4, you get 0/0, a discontinuity.
thanks guys @vf321 @mathmate