Hi

I'm having trouble finding the turning point(max & min) of this equation and also the stationary points and x-intercepts.

$\displaystyle y=9(x-1)^2-(x-1)^4$

P.S

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- Dec 2nd 2009, 02:05 PMPaymemoneyFinding the turning point
Hi

I'm having trouble finding the turning point(max & min) of this equation and also the stationary points and x-intercepts.

$\displaystyle y=9(x-1)^2-(x-1)^4$

P.S - Dec 2nd 2009, 02:24 PMBacterius
The x-intercepts are the points where the curve hits the x-axis. Do you mean you have trouble solving $\displaystyle f(x) = 0$ for $\displaystyle x$ ?

Think about using the derivative. When the derivative is $\displaystyle 0$ in a point, the function is very likely to turn round at this point (although sometimes it doesn't).

Steps :

- get the derivative $\displaystyle f'(x)$ of your function $\displaystyle f(x)$

- solve for $\displaystyle x$ with $\displaystyle f'(x) = 0$

- ensure each point found is indeed a turning point (you should know how to do this last step)

Does it make sense ? - Dec 2nd 2009, 05:29 PMPaymemoney
yeh it makes sense, thanks for the help