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

##### 1 Answer

#### 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]}