How Do You Find the X-Coordinate of Point Q on Curve C?

[thomas49th]

Hi

The curve C has equation y = f(x) and the point P(3,5) lies on C

Given that

f'(x) = 3x² - 8x + 6

The point Q also lies on C, and the tangent to C at Q is parallel to the tangent to C at P

Find the x-coordinate of Q

So

if there parallel the gradients are equal

but i can't really seem to get anywhere after that :(

 

[CompuChip]

Indeed, the gradients of y = f(x) at P and Q are equal. What is the value of that slope at P? Then, what other x gives the same slope?

 

[thomas49th]

You factorise the derative of f(x) which is f'(x)
giving you
(3x+1)(x-3)

x = 3 or x = -1/3

we already know x = 3 so the other is -1/3

i believe that is correct?

 

[HallsofIvy]

You factorise the derative of f(x) which is f'(x)
giving you
(3x+1)(x-3)

x = 3 or x = -1/3

we already know x = 3 so the other is -1/3

i believe that is correct?

Yes, that is correct but the way you phrased it confused me a bit! You did not factorize f '(x). You can easily calculate that the derivative at x= 3 is 9 so at the other point, you must have 3x^2- 8x+ 6= 9 which then gives 3x^3- 8x- 3= 0. That is what you factored, not the derivative of f.