No! yeKciM, your first answer was correct.
$\displaystyle \frac{x- 4}{x^2- x- 12}= \frac{x-4}{(x-4)(x+3)}$
is not defined for x= 4 or x= -3.
$\displaystyle \frac{x- 4}{(x- 4)(x+ 3)}= \frac{1}{x+3}$ only for $\displaystyle x\ne 4$.
If x= 4, $\displaystyle \frac{x- 4}{(x- 4)(x+ 3)}= \frac{0}{0}$ which is indeterminant while $\displaystyle \frac{1}{x+ 3}= \frac{1}{7}$.
well i think its just -3
reason why i think that is, if we look at function like
$\displaystyle f_{(x)}=\frac{x-4}{(x-4)(x+3)}$
it would mean that it's not defined in $\displaystyle x=4 , x=-3$ and that it has stationary point in $\displaystyle x=4$ but that isn't true ... it hasn't any stationary point's and it's defined in $\displaystyle x=4$ as u see from plot that i posted .... witch is actually function
$\displaystyle f_{(x)}=\frac {1}{x+3}$
lol... didn't have this much of polemics for such a simple function a long time
P.S. u shouldn't be just interested in "which choice is it a,b,c or d??"... it's more important to u to understand how and why does i say this, or HelsofIvy that, or skeeter... rather then be just interested in right answer... and soon have more problems (with same type) which u can't solve
No, it isn't. The graph of $\displaystyle y= \frac{x-4}{(x-4)(x+3)}$ looks like a hyperbola but with a hole at (4, 1/7). The graph of y= 1/(x+3) is the same hyperbola without the hole. If you plotted it with a calculator or a computer graphing program, your grid was probably too course, and it "jumped over" x= 4.
It can be crucially important in Calculus to distinguish things like $\displaystyle \frac{x-4}{(x-4)(x+3)}$ from $\displaystyle \frac{1}{x+3}$.