How do you identify all asymptotes or holes and intercepts for f(x)=(2x^2+3)/(x^2+6x+8)?

Dec 14, 2017

$\text{see explanation}$

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{ solve } {x}^{2} + 6 x + 8 = 0 \Rightarrow \left(x + 2\right) \left(x + 4\right) = 0$

$\Rightarrow x = - 4 \text{ and "x=-2" are the asymptotes}$

$\text{horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

$\text{divide terms on numerator/denominator by the highest}$
$\text{power of x that is } {x}^{2}$

$f \left(x\right) = \frac{\frac{2 {x}^{2}}{x} ^ 2 + \frac{3}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{6 x}{x} ^ 2 + \frac{8}{x} ^ 2} = \frac{2 + \frac{3}{x} ^ 2}{1 + \frac{6}{x} + \frac{8}{x} ^ 2}$

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{2 + 0}{1 + 0 + 0}$

$\Rightarrow y = 2 \text{ is the asymptote}$

$\text{holes occur if a common factor is eliminated from}$
$\text{the numerator/denominator. This is not the case here}$
$\text{hence there are no holes}$

$\textcolor{b l u e}{\text{Intercepts}}$

• " let x = 0 for y-intercept"

• " let y = 0 for x-intercepts"

$x = 0 \to y = \frac{3}{8} \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \to 2 {x}^{2} + 3 = 0 \Rightarrow {x}^{2} = - \frac{3}{2}$

$\text{this has no real solutions hence no x-intercepts}$
graph{(2x^2+3)/(x^2+6x+8) [-10, 10, -5, 5]}