How do you identify the oblique asymptote of $f (x) = \frac{2 x^{2} + 3 x + 8}{x + 3}$ $f(x) = (2x^2 + 3x + 8)/(x + 3)$ ?

Question

4307 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-05T23:04:48

$f (x) = \frac{2 x^{2} + 3 x + 8}{x + 3} = 2 x - 3 + \frac{17}{x + 3}$ $f(x) = (2x^2+3x+8)/(x+3) = 2x-3+17/(x+3)$

$\frac{17}{x + 3} \to 0$ $17/(x+3)->0$ as $x \to \pm \infty$ $x->+-oo$

So the oblique asymptote is the line $y = 2 x - 3$ $y = 2x-3$

$f (x) = \frac{2 x^{2} + 3 x + 8}{x + 3}$ $f(x) = (2x^2+3x+8)/(x+3)$

$= \frac{2 x^{2} + 6 x - 3 x - 9 + 17}{x + 3}$ $=(2x^2+6x-3x-9+17)/(x+3)$

$= \frac{2 x (x + 3) - 3 (x + 3) + 17}{x + 3}$ $=(2x(x+3)-3(x+3)+17)/(x+3)$

$= 2 x - 3 + \frac{17}{x + 3}$ $=2x-3+17/(x+3)$ with exclusion $x \neq - 3$ $x != -3$

graph{(y - (2x^2+3x+8)/(x+3))(y - (2x-3)) = 0 [-80, 80, -40, 40]}

How do you identify the oblique asymptote of f(x)=2x2+3x+8x+3f(x) = (2x^2 + 3x + 8)/(x + 3)?