How do you graph $y = \frac{2 x^{2} - 3}{x + 2}$ $y=(2x^2-3)/(x+2)$ using asymptotes, intercepts, end behavior?

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How do you graph $y = \frac{2 x^{2} - 3}{x + 2}$ $y=(2x^2-3)/(x+2)$ using asymptotes, intercepts, end behavior?

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Answer 1 · 2016-11-05T11:44:44

The vertical asymptote is $x = - 2$ $x=-2$
The oblique asymptote is $y = 2 x + 4$ $y=2x+4$

Explanation:

The vertical asymptote is $x = - 2$ $x=-2$
As the degree of numerator is $>$ $>$ the degree of denominator, we expect an oblique asymptote.
Therefore, we do a long division
$2 x^{2}$ $2x^2$ $a a a a$ $color(white)(aaaa)$ $- 3$ $-3$ $a a a a$ $color(white)(aaaa)$ $∣$ $∣$ x+2
$2 x^{2} + 4 x$ $2x^2+4x$ $a a$ $color(white)(aa)$ #color(white)(aaaa)∣#2x-4
$a$ $color(white)(a)$ $0 - 4 x - 3$ $0-4x-3$
$a$ $color(white)(a)$ $0 - 4 x - 8$ $0-4x-8$
$a a a a$ $color(white)(aaaa)$ $- 0 + 5$ $-0+5$
So, $\frac{2 x^{3} - 3}{x + 2} = 2 x - 4 + \frac{5}{x + 2}$ $(2x^3-3)/(x+2)=2x-4+5/(x+2)$
So the oblique asymptote is $y = 2 x - 4$ $y=2x-4$
Limit $y = 2 \frac{x^{2}}{x} = 2 x = \pm \infty$ $y=2x^2/x=2x=+-oo$
$x \to \pm \infty$ $x->+-oo$
graph{(y-(2x^2-3)/(x+2))(y-2x+4)=0 [-33.56, 31.36, -22.34, 10.15]}

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How do you graph $y = \frac{2 x^{2} - 3}{x + 2}$ $y=(2x^2-3)/(x+2)$ using asymptotes, intercepts, end behavior?

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