Hey guys, looking at this problem you may be wondering why it is not in the algebra section..but in the third part of the question it asks me to find the coordinates of the maximum point. I tried finding the t.p. by findind the dervative of the finction and fiding the zeros...but am not getting the right answer. I know Mr.Fantastic would advocate "working smart and not hard" by using the GDC. However, i dont want to that. Where am i going wrong...?

Q. y= a(x-b)(x-c)^2 y-intercept = -50

and graph crosses x-axis at x=1 and x=5

Find a, b and c

Easy enough, b=-1 and c=-5

I substituted x-solutions into the equation and solved for a.

Then it says that a point P is a local maximum on this graph between the solutions 0<x<5. Find the coordinates of P.

I opend out the equation:

f(x)= 2x^3 + 22x^2 + 70x + 50

f '(x) > 6x^2 + 44x + 70

2 [ 3x^2 + 22X + 35 ]

2 [ (3x(x+5) +7(x+5) ] = 0 Slope is zero at t.p.

Now im getting x = 0 and x = -5

as x-coordinate solutions for the turning point...doesnt make sense

... Should i take the second derivative maybe?

And when is it recomended to take the second derivative??

* Incidentally, GDC value is (2.53, 19)