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

##### 1 Answer

There is a vertical asymptote at

There is a horizontal asymptote as

x-intercepts when

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

Critical points are:

Maximum at

Minimum at

#### Explanation:

# y=(20x)/(x^2+1)-1/x #

# :. y=((20x^2)-(x^2+1))/(x(x^2+1)) #

# :. y=(19x^2-1)/(x(x^2+1)) #

**Vertical Asymptotes**

These will occur when the denominator is zero

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

So There is a vertical asymptote at

**Horizontal Asymptotes**

We need to examine the behaviour as

For large

So,

So There is a horizontal asymptote as

**Odd/Even Function**

Find y(-x);

# y(-x)=(19(-x)^2-1)/(-x((-x)^2+1)) #

# y(-x)=(19x^2-1)/(-x(x^2+1)) #

# y(-x)=-(19x^2-1)/(x(x^2+1)) #

# y(-x)=-y(x) #

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

**Intercepts**

# y=(19x^2-1)/(x(x^2+1)) => 19x^2-1 = 0 #

# :. x^2=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

**Summary**

There is a vertical asymptote at

There is a horizontal asymptote as

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

We can write;

# y=(19x^2-1)/(x^3+x) #

Using the quotient rule

# y=(19x^2-1)/(x^3+x) #

# dy/dx=((x^3+x)(d/dx(19x^2-1)) - (19x^2-1)(d/dx(x^3+x)) )/(x^3+x)^2 #

# :. dy/dx=((x^3+x)(38x) - (19x^2-1)(3x^2+1) )/(x^3+x)^2 #

# :. dy/dx=(38x^4+38x^2 - (57x^4+16x^2-1) )/(x^3+x)^2 #

# :. dy/dx=(38x^4+38x^2 - 57x^4-16x^2+1 )/(x^3+x)^2 #

# :. dy/dx=(-19x^4+22x^2 +1 )/(x^3+x)^2 #

# :. dy/dx=-((19x^4-22x^2 -1 ))/(x^3+x)^2 #

At extrema,

This is quadratic in

To solve this we use the quadratic formula

# x=(-b+-sqrt(b^2-4ac))/(2a)#

to find two solutions for

# x=+-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

By inspection (using a calculator with

When

Hence, Critical points are:

Maximum at

Minimum at