# How do you sketch the graph of f(x) = (x^2 - 3x + 9)/(x - 3)?

Apr 4, 2015

$f \left(x\right) = \frac{{x}^{2} - 3 x + 9}{x - 3} = \frac{{x}^{2} - 3 x}{x - 3} + \frac{9}{x - 3}$

$= \frac{x \left(x - 3\right)}{x - 3} + \frac{9}{x - 3} = x + \frac{9}{x - 3}$

So you actually wanna sketch $f \left(x\right) = x + \frac{9}{x - 3}$

Domain:
The domain of this function is all Real x except 3
because $\frac{9}{x - 3}$ would become unreal!

This brings us to Asymptotes:
They're two Asymptotes:

No 1: $y = x$ because as $x \to \infty$ $f \left(x\right) \to x$

No 2: $x = 3$ is a vertical asymptote because as $f \left(x\right) \to \infty$ $f \left(x\right) \to 3$

Next, turning points:
To find turning points (maxima and minima),
We let$y = f \left(x\right) \implies y = x + \frac{9}{x - 3}$

We make $x$ the subject of the equation,
$\implies y x - 3 y = {x}^{2} - 3 x + 9$
$\implies {x}^{2} - \left(3 + y\right) x + 9 + 3 y = 0$

Since all the values of $x$ are real, ${b}^{2} - 4 a c \ge 0$

$\implies {\left(3 + y\right)}^{2} - 4 \left(9 + 3 y\right) \ge 0$

$\implies {y}^{2} - 6 y - 27 \ge 0 \implies \left(y - 9\right) \left(y + 3\right) \ge 0$

Hence, $y \le - 3 \cup y \ge 9$

Putting the values of $y$ in $f \left(x\right)$ to get $x$
we have the points $\left(0 , - 3\right)$ maximum and $\left(6 , 9\right)$ as minimum

To find the intercepts :
$y$- intercept is when $x = 0$ we have $\left(0 , - 3\right)$
$x$- intercept is when $y = 0$ but $y$ cannot be $0$(see the range above)

Table of variation:

So now we're all set to sketch that graph;
Here's what i got: