Hi All SmilesAlways,
Since you are dividing by 5x-6,
$$5x-5\ne 0 \rightarrow x\ne 1$$
$$f/g=\frac{-20x^2+14x+12}{5x-6}\\\\
f/g=\frac{-2(10x^2+7x+6)}{5x-6}\\\\
f/g=\frac{-2(2x+1)(5x-6)}{5x-6}\\\\
f/g=-2(2x+1)$$
This is a line with a hole at (1,-6)
Domain All real x where x is not equal to 1
Range: All real y where y is not equal to -6
I am very tired so you need to check my working.
Sorry Melody, you may want to adjust some things;
$$5x-6 \neq 0$$
$$x \neq \frac{6}{5}$$
and
$$(2x+1)(5x-6) = 10x^2+5x-12x-6$$
which is
$$10x^2-7x-6$$
but since $$-20x^2+14x+6 = -2(10x^2-7x-6)$$
The rest does work.
Nevertheless the hole is at x= 6/5