Looking for the co-ordinates of the tangent parallel to this when x=0.6
2x^3-6x^2+10x+9
so far got - x=0.6 y=13.272 gradient = 4.96 (of original function)
how do i find the parallel tangent and then its co-ordinates.
Looking for the co-ordinates of the tangent parallel to this when x=0.6
2x^3-6x^2+10x+9
so far got - x=0.6 y=13.272 gradient = 4.96 (of original function)
how do i find the parallel tangent and then its co-ordinates.
I don't get this either. You are looking for "the co-ordinates of the tangent parallel to this" y =
2x^3-6x^2+10x+9 at the point (0.6, 13.272)? We can certainly get the tangent line to this curve at x = 0.6. Is this what you are looking for? What do you mean by "2nd tangent?" This is badly worded.
-Dan
I can only interpret this as meaning that you are looking for a value of x, other than .6 that has the same slope.
With $\displaystyle f(x)= 2x^3-6x^2+10x+9$, $\displaystyle f'(x)= 6x^2-12x+ 10$. When $\displaystyle x= .6$, that is
$\displaystyle f'(.6)= 6(.6)^2- 12(.6)+ 10= 4.96$
so now you are looking for another value of x that satisfies $\displaystyle f'(x)= 6x^2- 12x+10= 4.96$
That is the same as solving $\displaystyle 6x^2- 12x+ 5.04$ and that is especially easy to solve because you already know that x= .6 is a solution: divide $\displaystyle 6x^2- 12x+ 5.04$ by $\displaystyle x- .6$.