# 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?

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Aleord Share
Jun 16, 2018

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

#### Explanation:

Vertical asymptotes:
$- 3 {x}^{2} + 9 x - 6 = 0 \setminus \rightarrow - 3 \left({x}^{2} + 3 x - 2\right) = 0$
${x}^{2} + 3 x - 2 = 0$ where $a = 1$; $b = 3$ and $c = - 2$
$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a} = \setminus \frac{- 3 \setminus \pm \setminus \sqrt{{3}^{2} - 4 \cdot 1 \cdot \left(- 2\right)}}{2 \cdot 1} = \setminus \frac{- 3 \setminus \pm \setminus \sqrt{9 + 8}}{2} = \setminus \frac{- 3 \setminus \pm \setminus \sqrt{17}}{2}$
Are: ${x}_{1} = \setminus \frac{- 3 + \setminus \sqrt{17}}{2}$ and ${x}_{2} = \setminus \frac{- 3 - \setminus \sqrt{17}}{2}$.

Horizontal asymptote:
$L i {m}_{x \setminus \rightarrow + \setminus \infty} \setminus \frac{{x}^{3} - 6 {x}^{2} + 8 x}{- 3 {x}^{2} + 9 x - 6} = \setminus \frac{+ \setminus \infty}{- \setminus \infty}$
$L i {m}_{x \setminus \rightarrow + \setminus \infty} \setminus \frac{\setminus \frac{{x}^{3} - 6 {x}^{2} + 8 x}{{x}^{3}}}{\setminus \frac{- 3 {x}^{2} + 9 x - 6}{{x}^{3}}} =$

$L i {m}_{x \setminus \rightarrow + \setminus \infty} \setminus \frac{\setminus \frac{{x}^{3}}{{x}^{3}} - \setminus \frac{6 {x}^{2}}{{x}^{3}} + \setminus \frac{8 x}{{x}^{3}}}{\setminus \frac{- 3 {x}^{2}}{{x}^{3}} + \setminus \frac{9 x}{{x}^{3}} - \setminus \frac{6}{{x}^{3}}} =$

$L i {m}_{x \setminus \rightarrow + \setminus \infty} \setminus \frac{1 - \setminus \frac{6}{x} + \setminus \frac{8}{{x}^{2}}}{\setminus \frac{- 3}{x} + \setminus \frac{9}{{x}^{2}} - \setminus \frac{6}{{x}^{3}}} = \setminus \frac{1 - \setminus \frac{6}{+ \setminus \infty} + \setminus \frac{8}{{\left(+ \setminus \infty\right)}^{2}}}{\setminus \frac{- 3}{\left(+ \setminus \infty\right)} + \setminus \frac{9}{{\left(+ \setminus \infty\right)}^{2}} - \setminus \frac{6}{{\left(+ \setminus \infty\right)}^{3}}} = \setminus \frac{1 - 0 + 0}{0 + 0 - 0} = \setminus \frac{1}{0} = + \setminus \infty$

$L i {m}_{x \setminus \rightarrow - \setminus \infty} \setminus \frac{{x}^{3} - 6 {x}^{2} + 8 x}{- 3 {x}^{2} + 9 x - 6} = \setminus \frac{+ \setminus \infty}{- \setminus \infty}$ are $- \setminus \infty$

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

x intercept find the roots of ${x}^{3} - 6 {x}^{2} + 8 x$.
${x}^{3} - 6 {x}^{2} + 8 x = 0 \setminus \rightarrow x \left({x}^{2} - 6 x + 8\right) = 0$
$x = 0$ or ${x}^{2} - 6 x + 8 = 0$
where $a = 1$; $b = - 6$ and $c = 8$
$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a} = \setminus \frac{- \left(- 6\right) \setminus \pm \setminus \sqrt{{\left(- 6\right)}^{2} - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \setminus \frac{6 \setminus \pm \setminus \sqrt{36 - 32}}{2} = \setminus \frac{6 \setminus \pm \setminus \sqrt{4}}{2} = \setminus \frac{6 \setminus \pm 2}{2} =$
Are: ${x}_{1} = \setminus \frac{6 + 2}{2} = \setminus \frac{8}{2} = 4$ and ${x}_{2} = \setminus \frac{6 - 2}{2} = \setminus \frac{4}{2} = 2$.

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