1. ## Integral

A curve has its gradient functions as (2ax-b). It passes through the origin and (2,8) is its maximum point. Find a, b, and the equation of the curve.

I know a couple of things that its got something to do with the word maximum and make sure you understand and (2,8) as its maximum point.
I have no working to give as I have no idea where to begin.
Thanks

2. Originally Posted by Awsom Guy
A curve has its gradient functions as (2ax-b). It passes through the origin and (2,8) is its maximum point. Find a, b, and the equation of the curve.

I know a couple of things that its got something to do with the word maximum and make sure you understand and (2,8) as its maximum point.
I have no working to give as I have no idea where to begin.
Thanks
If $\displaystyle f'(x) = 2ax - b$ and $\displaystyle (2, 8)$ is the maximum, then $\displaystyle f'(2) = 0$.

Therefore $\displaystyle 2a(2) - b = 0$

$\displaystyle b = 4a$.

$\displaystyle f(x) = \int{2ax - b\,dx}$

$\displaystyle = ax^2 - bx + c$

And since the curve passes through $\displaystyle (0, 0)$ and $\displaystyle (2, 8)$ substitute these values into $\displaystyle f(x)$ and you will end up with some equations you can solve simultaneously for $\displaystyle a, b, c$.

3. Sorry I dont get that and whats the big F looking thing sorry that stuff I haven;t learnt.

4. Originally Posted by Awsom Guy
Sorry I dont get that and whats the big F looking thing sorry that stuff I haven;t learnt.
I haven't used a big F...

5. umm.. the big \int thing.

6. Originally Posted by Awsom Guy
umm.. the big \int thing.
It's the integral sign, and it stands for the indefinite integral, also known as the antiderivative.

So I'm saying take the antiderivative of that function.

7. ok thanks

8. but where does the dx come from.

9. Originally Posted by Awsom Guy
but where does the dx come from.
It's just notation.

It means take the antiderivative of the function with respect to the variable $\displaystyle x$.

It's important for when you learn the reverse chain rule.

10. so I get:
C=0 and:
8=4a^2-2b+C
How can I simultanously solve this.

11. Originally Posted by Awsom Guy
so I get:
C=0 and:
8=4a^2-2b+C
How can I simultanously solve this.
You know that $\displaystyle c = 0$ and $\displaystyle b = 4a$. Substitute them in.

12. does a equal to: (8)/(4a-8)
thanks

13. these are meant to be the answers:
a=-2
b=-8
y=-(2x^2)-8x

14. Originally Posted by Awsom Guy
these are meant to be the answers:
a=-2
b=-8
y=-(2x^2)-8x
You should be getting a quadratic equation in $\displaystyle a$. You'd need to factorise or use the Quadratic formula to evaluate $\displaystyle a$.

15. Sorry I am just not getting this, I'll just skip this question thanks anyways.

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