How do you graph #f(x)=(x^3-6x^2+8x)/(-3x^2+9x-6)# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Jun 16, 2018

Answer:

The vertical asymptotes are: #\frac{-3 \pm \sqrt{17}}{2}#.
The horizontal asymptote is 0
Y intercept is y=0
X intercept are: 0, 2 and 4

Explanation:

Vertical asymptotes:
#-3x^2+9x-6=0 \rightarrow -3(x^2 + 3x -2)=0#
#x^2+3x-2 =0# where #a=1#; #b=3# and #c=-2#
#x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} =\frac{-3 \pm \sqrt{3^2-4*1*(-2)}}{2*1} = \frac{-3 \pm \sqrt{9+8}}{2} = \frac{-3 \pm \sqrt{17}}{2} #
Are: #x_1=\frac{-3 + \sqrt{17}}{2}# and #x_2=\frac{-3 - \sqrt{17}}{2}#.

Horizontal asymptote:
#Lim_{x \rightarrow + \infty} \frac{x^3-6x^2+8x}{-3x^2+9x-6} = \frac{+\infty}{- \infty}#
#Lim_{x \rightarrow + \infty} \frac{\frac{x^3-6x^2+8x}{x^3}}{\frac{-3x^2+9x-6}{x^3}} = #

#Lim_{x \rightarrow + \infty} \frac{\frac{x^3}{x^3}-\frac{6x^2}{x^3}+\frac{8x}{x^3}}{\frac{-3x^2}{x^3}+\frac{9x}{x^3}-\frac{6}{x^3}} = #

# Lim_{x \rightarrow + \infty} \frac{1-\frac{6}{x}+\frac{8}{x^2}}{\frac{-3}{x}+\frac{9}{x^2}-\frac{6}{x^3}} = \frac{1-\frac{6}{+\infty}+\frac{8}{(+\infty)^2}}{\frac{-3}{(+\infty)}+\frac{9}{(+\infty)^2}-\frac{6}{(+\infty)^3}} = \frac{1-0+0}{0+0-0} = \frac{1}{0} = + \infty#

#Lim_{x \rightarrow - \infty} \frac{x^3-6x^2+8x}{-3x^2+9x-6} = \frac{+\infty}{- \infty}# are #-\infty#

y intercept: evaluate x=0 in the function
#\frac{0^3-6*0^2+8*0}{-3*0^2+9*0-6} = \frac{0}{-6} =0#

x intercept find the roots of #x^3-6x^2+8x#.
#x^3-6x^2+8x = 0 \rightarrow x(x^2-6x+8) = 0#
#x =0# or #x^2-6x+8 = 0#
where #a=1#; #b=-6# and #c=8#
#x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} =\frac{-(-6) \pm \sqrt{(-6)^2-4*1*8}}{2*1} = \frac{6 \pm \sqrt{36-32}}{2} = \frac{6 \pm \sqrt{4}}{2} =\frac{6 \pm 2}{2} = #
Are: #x_1=\frac{6 + 2}{2} = \frac{8}{2} = 4# and #x_2=\frac{6 - 2}{2} = \frac{4}{2} = 2#.