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?
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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 $\displaystyle x = \frac{8}{3}$.
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 $\displaystyle x=\frac{8}{3}$ Thanks for your help.
You're told that $\displaystyle y = 4x^2 - x^3$. Now let $\displaystyle x = \frac{8}{3}$ and find $\displaystyle y$.
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.
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