We have f(x) = g(x) / (h(x)) = (2x^3 -x^2 - 2x +1) / (x^2+3x+2)f(x)=g(x)h(x)=2x3−x2−2x+1x2+3x+2.
Function h(x)h(x) quite easily factorises down to (x+1)(x+2)(x+1)(x+2).
Divide g(x)g(x) by x+1x+1 to factorise the cubic, and it can be found through polynomial division that g(x) = (x+1)(2x^2-3x+1)g(x)=(x+1)(2x2−3x+1), the latter quadratic of which factorises down to (2x-1)(x-1)(2x−1)(x−1).
Hence f(x) = ((2x-1)(x-1)(x+1)) / ((x+1)(x+2)) = ((2x-1)(x-1)) / (x+2)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)2x2−3x+1x+2, it can be found that f(x) = 2x - 7 + 15/(x+2)f(x)=2x−7+15x+2.
Clearly xx cannot equal -2−2 as it leaves the fraction undefined, so there is an asymptote at x=-2x=−2.
Secondly f(x)f(x) cannot equal 2x-72x−7, as this would suggest that 15/(x+2) = 0 rArr 15 = 015x+2=0⇒15=0, so there is an oblique asymptote at y=2x-7y=2x−7.
To solve for xx-intercept(s) return to the factorised function, from which it can be seen easily that for f(x)f(x) to equal zero, (2x-1)(x-1) = 0(2x−1)(x−1)=0, therefore x=1/2x=12 or 11 for the xx-intercepts.
Let x = 0x=0 to find yy-intercepts. Thus f(x) = ((-1)(-1))/2f(x)=(−1)(−1)2, therefore y=1/2y=12.
Lastly, the 'holes'. Two x+1x+1 expressions were cancelled early on, hence there is a hole at x=-1x=−1. So, a discontinuity is found at (-1, f(-1)) -= (-1, 6)(−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!
