How do you identify all asymptotes or holes for h(x)=(2x^2+9x+2)/(2x+3)?

Aug 30, 2017

$\text{vertical asymptote at } x = - \frac{3}{2}$
$\text{oblique asymptote } y = x + 3$

Explanation:

The denominator of h(x) cannot be zero as tis would make h(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

$\text{solve "2x+3=0rArrx=-3/2" is the asymptote}$

Horizontal asymptotes occur when the degree of the numerator $\le$ degree of the denominator. This is not the case here hence there are no horizontal asymptotes.

Oblique (slant ) asymptotes occur when the degree of the numerator $>$ degree of the denominator. This is the case here hence there is an oblique asymptote.

$\text{dividing out}$

$x \left(2 x + 3\right) + 3 \left(2 x + 3\right) - 7$

$\Rightarrow h \left(x\right) = \frac{2 {x}^{2} + 9 x + 2}{2 x + 3} = x + 3 - \frac{7}{2 x + 3}$

$\text{as } x \to \pm \infty , h \left(x\right) \to x + 3$

$\Rightarrow y = x + 3 \text{ is the asymptote}$

Holes occur when there is a cancellation of a common factor on the numerator/denominator. This is not the case here hence there are no holes.
graph{(2x^2+9x+2)/(2x+3) [-10, 10, -5, 5]}