Let
u=2x^3-1
v=x^2-2x
Then
f(x)=u/v
Using the quotient rule, which can be found in all good google searches:
f'(x)=(u'v-uv')/v^2
u'=6x
v'=2x-2
f'(x)=([6x(x^2-2x)]-[(2x^3-1)(2x-2)])/(x^2-2x)^2
Soz, I cannot be bothered simplifying this...
f'(x) is the gradient function, substituting in any x value will give us the gradient of the tangent at that point.
So let's sub in x=3:
f'(3)=([6*3(3^2-2*3)]-[(2*3^3-1)(2*3-2)])/(3^2-2*3)^2
rArrf'(3)=([18(9-6)]-[(54-1)(6-2)])/(9-6)^2=(18*3-53*4)/9=-158/9
Now we want the equation of the tangent, which is a straight line of the form:
y=mx+c
We know m=-158/9 but we need to solve for c. We need a point on the line to solve it, so sub in x=3 into the original equation to find the y value:
f(3)=(2*3^3-1)/(3^2-2*3)=(54-1)/(9-6)=53/3
So the coordinates of our point on the line are (3,53/3). Sub this point into the straight line equation to solve for c:
y=-158/9*x+c
rArr53/3=-158/9*3+c
rArr53/3=-158/3+c
rArrc=53/3+158/3=211/3
So the equation of the tangent is:
y=-158/9*x+211/3