How do you graph f(x)=(2x^3-x^2-2x+1)/(x^2+3x+2) using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Oct 8, 2017

See below for more details about holes, intercepts, asymptotes, and stationary points too!

Explanation:

We have f(x) = g(x) / (h(x)) = (2x^3 -x^2 - 2x +1) / (x^2+3x+2).

Function h(x) quite easily factorises down to (x+1)(x+2).

Divide g(x) by x+1 to factorise the cubic, and it can be found through polynomial division that g(x) = (x+1)(2x^2-3x+1), the latter quadratic of which factorises down to (2x-1)(x-1).

Hence f(x) = ((2x-1)(x-1)(x+1)) / ((x+1)(x+2)) = ((2x-1)(x-1)) / (x+2)

By further polynomial division, that is (2x^2-3x+1)/(x+2), it can be found that f(x) = 2x - 7 + 15/(x+2).

Clearly x cannot equal -2 as it leaves the fraction undefined, so there is an asymptote at x=-2.

Secondly f(x) cannot equal 2x-7, as this would suggest that 15/(x+2) = 0 rArr 15 = 0, so there is an oblique asymptote at y=2x-7.

To solve for x-intercept(s) return to the factorised function, from which it can be seen easily that for f(x) to equal zero, (2x-1)(x-1) = 0, therefore x=1/2 or 1 for the x-intercepts.

Let x = 0 to find y-intercepts. Thus f(x) = ((-1)(-1))/2, therefore y=1/2.

Lastly, the 'holes'. Two x+1 expressions were cancelled early on, hence there is a hole at x=-1. So, a discontinuity is found at (-1, f(-1)) -= (-1, 6).

To find stationary points, calculate f'(x) and let it equal zero:
f'(x) = x - 15/((x+2)^2) = 0.
:. x(x+2)^2 = 15
:. x^3 + 4x^2 + 4x -15 = 0
So you end up with a stationary point at approximately (1.34, f(1.34)).

Hope this helps!