1. ## Turning Points

Sketch this curve on a graph

$\displaystyle y=4x^2-x^3$

I differentiated and got

$\displaystyle \frac{dy}{dx}=8x-3x^2$

At TP dy by dx = o

So $\displaystyle 8x-3x^2=0$

Now where do I go?

2. Originally Posted by Ryannnn
Sketch this curve on a graph

$\displaystyle y=4x^2-x^3$

I differentiated and got

$\displaystyle \frac{dy}{dx}=8x-3x^2$

At TP dy by dx = o

So $\displaystyle 8x-3x^2=0$

Now where do I go?
Factorise.

$\displaystyle 8x - 3x^2 = 0$

$\displaystyle x(8 - 3x) = 0$

$\displaystyle x = 0$ or $\displaystyle x = \frac{8}{3}$.

3. Originally Posted by Prove It
Factorise.

$\displaystyle 8x - 3x^2 = 0$

$\displaystyle x(8 - 3x) = 0$

$\displaystyle x = 0$ or $\displaystyle x = \frac{8}{3}$.
So when $\displaystyle x=0$

$\displaystyle y=0$

But how do you find $\displaystyle y$ when $\displaystyle x=\frac{8}{3}$

4. You're told that

$\displaystyle y = 4x^2 - x^3$.

Now let $\displaystyle x = \frac{8}{3}$ and find $\displaystyle y$.

5. Originally Posted by Prove It
You're told that

$\displaystyle y = 4x^2 - x^3$.

Now let $\displaystyle x = \frac{8}{3}$ and find $\displaystyle y$.
So...

When $\displaystyle x=\frac{8}{3}$

$\displaystyle y=\frac{256}{9}-\frac{512}{27}$

$\displaystyle y=\frac{256}{27}$

$\displaystyle y=9\frac{13}{27}$

Sorry for going on abit.