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

Sep 11, 2015

See explanation section.

#### Explanation:

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

Domain: $\mathbb{R} \text{ - } \left\{\frac{9}{2}\right\}$

$x$-intercept: $\frac{3}{2}$
$y$ intercept $- \frac{1}{27}$

Symmetry: None

Vertical Asymptote: $x = \frac{9}{2}$
Horizontal Asymptote: $y = 0$ (a.k.a. the $x$-axis)

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

Sign analysis of $f '$ shows that:
$f$ is decreasing on $\left(- \infty , - \frac{3}{2}\right)$ and on $\left(\frac{9}{2} , \infty\right)$ and
$f$ is increasing on $\left(- \frac{3}{2} , \frac{9}{2}\right)$

Local Minimum $- \frac{1}{24}$ (at $x = - \frac{3}{2}$)

$f ' ' \left(x\right) = \frac{8 \left(2 x + 9\right)}{2 x - 9} ^ 4$
Sign analysis of $f ' '$ shows that:
$f$ is concave down on $\left(- \infty , - \frac{9}{2}\right)$ and
$f$ is concave up on $\left(- \frac{9}{2} , \frac{9}{2}\right)$ and on $\left(\frac{9}{2} , \infty\right)$

$\left(- \frac{9}{2} , - \frac{1}{27}\right)$ is the only inflection point.

The graph looks like this:

graph{y=(2x-3) / (2 x -9)^2 [-6.16, 11.62, -2.91, 5.985]}

(You can scroll in and out using a mouse wheel and also drag the graph around to see features more clearly.)