Question

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

Answer 1

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

Explanation:

`f(x) = (6x^2+10x-3)/(2x+2) = 3x+2-7/(2x+2)`

To find the axis intercept points solve `f(x) = 0` and `f(0)`:
`f(x) = 0`
`6x^2+10x-3 = 0`
`x = (-5 pm sqrt(43))/6`

`f(0) = -3/2`

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

No horizontal asymptotes since:
`lim_(x rarr pm oo} f(x) = pm oo`

However:
`g(x) = 3x+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):
`(df)/dx ne 0, x in RR`
So there are no stationary points.

Points of inflection (second derivative is equal to zero):
`(d^2f)/dx^2 ne 0, x in RR`
So there are no points of inflection.

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