1. ## Cubic equation

I've been stuck on this problem for a while, after starting C1 AS level maths a few weeks ago. Here is the question:

Curve C => y = ax³ - x² + x - 5, a is some constant.
Point P => (-1/2, y)
Tangent to C at P is parallel to y = (7x - 1)/2

Find a.

Any help will be appreciated, thanks.

2. ## Re: Cubic equation

Originally Posted by b150
I've been stuck on this problem for a while, after starting C1 AS level maths a few weeks ago. Here is the question:

Curve C => y = ax³ - x² + x - 5, a is some constant.
Point P => (-1/2, y)
Tangent to C at P is parallel to y = (7x - 1)/2

Find a.

Any help will be appreciated, thanks.
Parellel to y = (7/2)x - 1/2 means the slope of the tangent (at x = -1/2) is 7/2
So if f(x) = ax^3 -x^2 + x - 5, then f ' (-1/2) = ...

Can you take it from here?

3. ## Re: Cubic equation

Originally Posted by b150
I've been stuck on this problem for a while, after starting C1 AS level maths a few weeks ago. Here is the question:

Curve C => y = ax³ - x² + x - 5, a is some constant.
Point P => (-1/2, y)
Tangent to C at P is parallel to y = (7x - 1)/2

Find a.

Any help will be appreciated, thanks.
The tangent of C is $\displaystyle f'\left(-\dfrac{1}{2}\right)$ and since parallel lines have the same gradient $\displaystyle f'\left(-\dfrac{1}{2}\right) = \dfrac{7}{2}$

Solve for a

edit: too slow

4. ## Re: Cubic equation

This looks like a calculus question.

Do you know how to find the derivative of this cubic function?

Do you know what a derivative tells about a function?

5. ## Re: Cubic equation

Originally Posted by e^(i*pi)
...

edit: too slow
...but much better looking!

6. ## Re: Cubic equation

haha I realised how to do it like 5 minutes after I posted :P Thanks anyway.

7. ## Re: Cubic equation

Originally Posted by TheChaz
...but much better looking!
I don't see how you can say that e^(i*pi) is much better looking ! He doesn't even show a picture of himself.

... Oh! You mean his post was much better looking? Yes, it does look good -- LaTeX does that.