Question

How do you graph `f(x)=(3x^2-2x)/(x-1)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

see explanation.

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

`"solve " x-1=0rArrx=1" is the asymptote"`

Since the degree of the numerator > degree of the denominator there will be an oblique asymptote but no horizontal asymptote.

To obtain oblique asymptote, divide numerator by denominator.

`" Consider the numerator"`

`color(red)(3x)(x-1)color(magenta)(+3x)-2x`

`=color(red)(3x)(x-1)color(red)(+1)(x-1)color(magenta)(+1)`

`rArr" oblique asymptote is " y=3x+1`

`color(blue)"Intercepts"`

`x=0toy=0to(0,0)larrcolor(red)" y-intercept"`

`y=0to3x^2-2x=0tox(3x-2)=0`

`rArrx=0,x=2/3larrcolor(red)" x-intercepts"`
graph{(3x^2-2x)/(x-1) [-20, 20, -10, 10]}