Answer
The function is discontinuous because $f(3) = 6 \ne 7$.
Work Step by Step
For this function to be continuous by definition: $\lim\limits_{x \to 3} f(x) = f(3)$
$ \lim\limits_{x \to 3} \frac{2x^{2}-5x-3}{x-3} = $
$ \lim\limits_{x \to 3} \frac{(2x+1)(x-3)}{x-3} =$
Cancel out $(x-3)$
$ \lim\limits_{x \to 3} (2x+1) =$
$ \lim\limits_{x \to 3} (2(3)+1) = 7$
The value of $f(3) = 6$, but we discovered that $ \lim\limits_{x \to 3} (2(3)+1) = 7$ so the function is discontinuous.
To draw the graph we draw the function of $f(x) = (2x+1)$ and $(3,6)$.
