1. ## Trig Problem

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)

2. Originally Posted by BobBali
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)
The problem stems from writing b=-1 and c=-5 instead of the correct

b = 1
c = 5

The funny thing is that you must have switched the signs again to get the correct a = 2, because using b=-1 and c=-5 would lead to the incorrect a = -2.

After expansion you should get

$\displaystyle f(x)= 2x^3 \color{red}-\color{black} 22x^2 + 70x \color{red}-\color{black} 50$

After that it should be smooth sailing.

Edit: Ah, missed that b = 5, c = 1 also works. Silly of me, somehow thought that using those values would force a positive y-intercept. I think I was influenced by the wording of the question which seemed to imply there is a unique solution, so I didn't bother questioning once a correct solution was found.

3. Originally Posted by BobBali
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.
first of all, the graph does not "cross" the x-axis twice ... there are two zeros, but one of them has multiplicity two, meaning the curve just "touches" (not crosses) the x-axis at that zero.

two possible curves that fit this description ...

$\displaystyle y = 2(x-1)(x-5)^2$

and

$\displaystyle y = 10(x-5)(x-1)^2$

with two possible maximums ... the first curve yields the stated maximum.

4. Agh, i see where i went wrong.

2[3x(x-5) -7(x-5)] = 0

I forgot to put factorized terms together

2[(3x -7) (x-5)] = 0

Now, 3x-7 = 0 and x = 5

x = 7/3 = 2.33 Works out! Thanx