# How do you find intercepts, extrema, points of inflections, asymptotes and graph y=(20x)/(x^2+1)-1/x?

Nov 14, 2016

There is a vertical asymptote at $x = 0$
There is a horizontal asymptote as $x \rightarrow \pm \infty$ along the $x$-axis
x-intercepts when $x = \pm \frac{1}{\sqrt{19}}$
y(x) is an odd function, and it is symmetric about the origin O

Critical points are:
Maximum at $\left(1.096 , 9.046\right)$
Minimum at $\left(- 1.096 , - 9.046\right)$

#### Explanation:

$y = \frac{20 x}{{x}^{2} + 1} - \frac{1}{x}$
$\therefore y = \frac{\left(20 {x}^{2}\right) - \left({x}^{2} + 1\right)}{x \left({x}^{2} + 1\right)}$
$\therefore y = \frac{19 {x}^{2} - 1}{x \left({x}^{2} + 1\right)}$

Vertical Asymptotes
These will occur when the denominator is zero

 :. x(x^2+1)) =0 => x=0

So There is a vertical asymptote at $x = 0$

Horizontal Asymptotes
We need to examine the behaviour as $x \rightarrow \pm \infty$.
For large $x$,
 19x^2-1 ~ x^2
 x^2+1 ~ x^2
So,  (19x^2-1)/(x(x^2+1)) ~ 19x^2/x^3 ~ 1/x ->0 as $x \rightarrow \pm \infty$
So There is a horizontal asymptote as $x \rightarrow \pm \infty$ along the $x$-axis

Odd/Even Function
Find y(-x);

$y \left(- x\right) = \frac{19 {\left(- x\right)}^{2} - 1}{- x \left({\left(- x\right)}^{2} + 1\right)}$
$y \left(- x\right) = \frac{19 {x}^{2} - 1}{- x \left({x}^{2} + 1\right)}$
$y \left(- x\right) = - \frac{19 {x}^{2} - 1}{x \left({x}^{2} + 1\right)}$
$y \left(- x\right) = - y \left(x\right)$

Hence y(x) is an odd function, and it is symmetric about the origin O

Intercepts

$y = \frac{19 {x}^{2} - 1}{x \left({x}^{2} + 1\right)} \implies 19 {x}^{2} - 1 = 0$
$\therefore {x}^{2} = \frac{1}{19}$
 :. x^2=+-1/sqrt19 ~~~ +- 0.229

We have already established that x=0 is a vertical asymptotes so there are not y-axis intercepts.
Hence x-intercepts when $x = \pm \frac{1}{\sqrt{19}}$

Summary
There is a vertical asymptote at $x = 0$
There is a horizontal asymptote as $x \rightarrow \pm \infty$
y(x) is an odd function, and it is symmetric about the origin O

Extrema
Because of symmetry we only need to identify extra for $x \ge 0$
We can write;

$y = \frac{19 {x}^{2} - 1}{{x}^{3} + x}$

Using the quotient rule

$y = \frac{19 {x}^{2} - 1}{{x}^{3} + x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left({x}^{3} + x\right) \left(\frac{d}{\mathrm{dx}} \left(19 {x}^{2} - 1\right)\right) - \left(19 {x}^{2} - 1\right) \left(\frac{d}{\mathrm{dx}} \left({x}^{3} + x\right)\right)}{{x}^{3} + x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left({x}^{3} + x\right) \left(38 x\right) - \left(19 {x}^{2} - 1\right) \left(3 {x}^{2} + 1\right)}{{x}^{3} + x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{38 {x}^{4} + 38 {x}^{2} - \left(57 {x}^{4} + 16 {x}^{2} - 1\right)}{{x}^{3} + x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{38 {x}^{4} + 38 {x}^{2} - 57 {x}^{4} - 16 {x}^{2} + 1}{{x}^{3} + x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 19 {x}^{4} + 22 {x}^{2} + 1}{{x}^{3} + x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\left(19 {x}^{4} - 22 {x}^{2} - 1\right)}{{x}^{3} + x} ^ 2$

At extrema, $\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 19 {x}^{4} - 22 {x}^{2} - 1 = 0$
This is quadratic in ${x}^{2}$ and can be solved, which I will return to!
To solve this we use the quadratic formula

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

to find two solutions for ${x}^{2}$, and then take their square roots, which yields two real solutions;

$x = \pm 1.096$ (3dp)

Due to the symmetry about O, we only need to examine one of these critical points as the other will be the opposite. Let's look at $x = 1.096$

By inspection (using a calculator with $x = 1.096 \pm \epsilon$, where $\epsilon$ is small we can see this point corresponds to a maximum, and so $x = - 1.096$ is a minimum. We could also investigate using the second derivative test, but due to the complexity a simple computed test is easier.

When $x = 1.096 \implies y = 9.046$ (3dp)

Hence, Critical points are:
Maximum at $\left(1.096 , 9.046\right)$
Minimum at $\left(- 1.096 , - 9.046\right)$