# How do you find intercepts, extrema, points of inflections, asymptotes and graph f(x)=(2x^2-5x+5)/(x-2)?

Sep 20, 2017

Two asymptotes; two $x$-intercepts; one $y$-intercept; and two stationary points.

#### Explanation:

First of all do polynomial division to get your function in the nice form of:
$f \left(x\right) = 2 x + 1 + \frac{3}{x - 2}$.

From this it is fairly clear where the asymptotes are, as $x$ cannot equal $2$ as $f \left(x\right)$ would then be undefined. Similarly $f \left(x\right)$ cannot equal $2 x + 1$, as this would mean that $\frac{3}{x - 2}$ equals $0$, which it cannot as it implies $3 = 0$.

Then differentiate this function, applying the chain rule and product rules where necessary.

The derivative of $2 x + 1$ is of course $2$.

The derivative of $\frac{3}{x - 2}$ is done by letting $y = 3 {\left(x - 2\right)}^{-} 1$.

First apply the chain rule: say $t = x - 2$ and $w = {t}^{-} 1$.
$\therefore \frac{\mathrm{dt}}{\mathrm{dx}} = 1$ and $\frac{\mathrm{dw}}{\mathrm{dt}} = - {t}^{-} 2$.
$\therefore \frac{\mathrm{dw}}{\mathrm{dx}} = - {t}^{-} 2 = - {\left(x - 2\right)}^{-} 2$

The product rule isn't really necessary as it's pretty clear you will end up with:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{x - 2} ^ 2$.

$\therefore f ' \left(x\right) = 2 - \frac{3}{x - 2} ^ 2$.

Letting $f ' \left(x\right) = 0$ next.

$\therefore 3 = 2 {\left(x - 2\right)}^{2}$

Simplify it down to arrive at $2 {x}^{2} - 8 x + 8 = 0$, which by the quadratic formula leaves us with: $x = \frac{4 - \sqrt{6}}{2}$ or $x = \frac{\sqrt{6} + 4}{2}$.

Solving for $f \left(\frac{4 - \sqrt{6}}{2}\right)$ and $f \left(\frac{\sqrt{6} + 4}{2}\right)$ yields $5 - \sqrt{6}$ and $2 \sqrt{6} + 5$, which are maximum and minimum respectively.

Finally calculate your intercepts, first by letting $f \left(x\right) = 0$

$\therefore - 3 = \left(2 x + 1\right) \left(x - 2\right)$

$\therefore 2 {x}^{2} - 3 x + 1 = 0$

$\therefore x = \frac{1}{2}$ or $x = 1$.

Finally calculate $f \left(0\right)$ which is simply $1 - \frac{3}{2}$, therefore the $y$-intercept is where $y = - \frac{1}{2}$.