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