Question

f(x)=(x^2-3x-18)/x+3

Answer 1

What is the question?

Answer 2

find the restriction on x
factor numerator
cancel factors
graph the linear term with point of discontinuity at said restriction

Answer 3

\[f(x)=\frac{ x^{2}-3x-18 }{ x+3 }\]
From the start, in this form we can see that x can't equal -3. This would yield a solution of:
\[x=-3\]\[f(-3) \rightarrow \frac{ 0 }{ 0 }\]
Which is an indeterminate form. Factor the numerator and cancel:
\[f(x)=\frac{ (x+3)(x-6) }{ x+3 }=x-6\]
The graph is attached.

Answer 4

sweeeeeeet. now I wonder if he got that