How do you find the asymptotes for $f (x) = \frac{3 x^{2} - 2 x - 1}{x + 4}$ $f (x ) = (3x^2 - 2x - 1) / (x + 4 )$ ?

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How do you find the asymptotes for $f (x) = \frac{3 x^{2} - 2 x - 1}{x + 4}$ $f (x ) = (3x^2 - 2x - 1) / (x + 4 )$ ?

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Answer 1 · 2017-11-16T18:03:39

Vertical asymptote $x = - 4$ $x=-4$

Explanation:

$3 x^{2} - 2 x - 1$ $3x^2-2x-1$
$x_{1, 2} = \frac{2 \pm \sqrt{4 - 4 \cdot 3 \cdot (- 1)}}{2 \cdot 3} = \frac{2 \pm 4}{6}$ $x_(1,2)=(2+-sqrt(4-4*3*(-1)))/(2*3)=(2+-4)/6$
$\Rightarrow$ $=>$
$x_{1} = 1$ $x_1=1$
$x_{2} = - \frac{1}{3}$ $x_2=-1/3$
$\Rightarrow$ $=>$
$f_{(x)} = \frac{(x - 1) (3 x + 1)}{x + 4}$ $f_((x))=((x-1)(3x+1))/(x+4)$
$\Rightarrow$ $=>$
Vertical asymptote $x = - 4$ $x=-4$

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How do you find the asymptotes for $f (x) = \frac{3 x^{2} - 2 x - 1}{x + 4}$ $f (x ) = (3x^2 - 2x - 1) / (x + 4 )$ ?

1 Answer

Explanation:

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How do you find the asymptotes for f (x ) = (3x^2 - 2x - 1) / (x + 4 )f(x)=3x2−2x−1x+4f (x ) = (3x^2 - 2x - 1) / (x + 4 )?

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Explanation:

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How do you find the asymptotes for $f (x) = \frac{3 x^{2} - 2 x - 1}{x + 4}$ $f (x ) = (3x^2 - 2x - 1) / (x + 4 )$ ?