How do you graph (6x^2 + 10x - 3) /( 2x + 2)?

Aug 12, 2015

When plotting graphs, I usually find the axis intercept points, the asymptotes, the stationary points and the points of inflection.

Explanation:

$f \left(x\right) = \frac{6 {x}^{2} + 10 x - 3}{2 x + 2} = 3 x + 2 - \frac{7}{2 x + 2}$

To find the axis intercept points solve $f \left(x\right) = 0$ and $f \left(0\right)$:
$f \left(x\right) = 0$
$6 {x}^{2} + 10 x - 3 = 0$
$x = \frac{- 5 \pm \sqrt{43}}{6}$

$f \left(0\right) = - \frac{3}{2}$

Vertical asymptotes (denominator of f(x) = 0):
$2 x + 2 = 0$
$x = - 1$
 lim_(x rarr -1^(+-)) f(x) = ""_+^(-) oo

No horizontal asymptotes since:
${\lim}_{x \rightarrow \pm \infty} f \left(x\right) = \pm \infty$

However:
$g \left(x\right) = 3 x + 2$ is an asymptote since  lim_(x rarr pm oo) -7/(2x+2) = 0^(""^(""_+^-))

 lim_(x rarr pm oo} f(x) rArr lim_(x rarr pm oo} g(x)^(""^(""_+^-))

Stationary points (first derivative is equal to zero):
$\frac{\mathrm{df}}{\mathrm{dx}} \ne 0 , x \in \mathbb{R}$
So there are no stationary points.

Points of inflection (second derivative is equal to zero):
$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 \ne 0 , x \in \mathbb{R}$
So there are no points of inflection.

graph{(6x^2+10x-3)/(2x+2) [-5, 5, -25, 25]}