How do you graph $f (x) = \frac{3 x^{2} + 2}{x + 1}$ $f(x)=(3x^2+2)/(x+1)$ using holes, vertical and horizontal asymptotes, x and y intercepts?

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Answer 1 · 2018-02-18T02:16:59

V.A $x = - 1$ $x=-1$
H.A none
S.A $y = 3 x - 3$ $y=3x-3$
no $x$ $x$ -intercept
$y$ $y$ -intercept is $(0, 2)$ $(0,2)$
no holes

V.A are the zeros (undefined points) of the denominator
$x + 1 = 0$ $x+1=0$
$x = - 1$ $x= -1$
vertical asymptotes are $x = - 1$ $x= -1$
H.A if the degrees of the numerator is equal to the degree of denominator,
if the numerators degree > 1 + degree of denominator, there is a slant asymptote
degree of numerator is 2, degree of denominator is 1
$y = m x + b$ $y=mx+b$
$\frac{3 x^{2} + 2}{x + 1}$ $(3x^2+2)/(x+1)$
$= 3 x + \frac{- 3 x + 2}{x + 1}$ $=3x+(-3x+2)/(x+1)$
$= 3 x - 3 + \frac{5}{x + 1}$ $=3x-3+(5)/(x+1)$
$y = 3 x - 3$ $y=3x-3$
slant asymptote is $y = 3 x - 3$ $y=3x-3$
$x$ $x$ -intercept is a point on the graph where $y = 0$ $y=0$
$\frac{3 x^{2} + 2}{x + 1} = 0$ $(3x^2+2)/(x+1)=0$ no solution for $x inRR$
no $x$ -intercept
$y$ -intercept is the point on the graph where $x=0$
$y=(3 * 0^2+2)/(0+1)$
$y=2$
$y$ -intercept is $(0,2)$
no holes because the denominator doesn't cancel out

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How do you graph f(x)=(3x^2+2)/(x+1)f(x)=3x2+2x+1f(x)=(3x^2+2)/(x+1) using holes, vertical and horizontal asymptotes, x and y intercepts?