Question

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

Answer 1

`"see explanation"`

Explanation:

`"factorise the numerator/denominator"`

`f(x)=(cancel(x)(x+3))/(cancel(x)(x-1))=(x+3)/(x-1)`

`"the cancelling of the factor x indicates a hole at x = 0"`

`"substitute x = 0 into "f(x)=(x+3)/(x-1)`

`rArrf(0)=3/(-1)=-3`

`rArr"there is a hole at "(0,-3)`

`"the graph of "(x+3)/(x-1)" is the same as the graph of"`

`(x^2+3x)/(x^2-x)" but without the hole"`

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"`

`"horizontal asymptotes occur as"`

`lim_(xto+-oo),f(x)toc" (a constant)"`

`"divide terms on numerator/denominator by x"`

`f(x)=(x/x+3/x)/(x/x-1/x)=(1+3/x)/(1-1/x)`

`"as "xto+-oo,f(x)to(1+0)/(1-0)`

`rArry=1" is the asymptote"`

`color(blue)"for intercepts"`

`• " let x = 0 for y-intercept"`

`• " let y = 0 for x-intercepts"`

`f(0)=-3larrcolor(red)"y-intercept"`

`y=0tox+3=0rArrx=-3larrcolor(red)"x-intercept"`
graph{(x+3)/(x-1) [-10, 10, -5, 5]}