#1
ive to find the highest point it reaches, but i keep coming up with crazy numbers
please walk me through it

y= -6/289(x-205)^2+119
Quote by WickedBeast666
Noooooooooooo how could this be! he at all the chocolate in the box
Oh well, now the empty shell of what used to be chocolaty goodness can contain a tasty guitar circuit.


speaking of my homemade pedal
#4
Quote by fosho
ive to find the highest point it reaches, but i keep coming up with crazy numbers
please walk me through it

y= -6/289(x-205)^2+119


119. Any number squared is 0 or greater. As overything but the +119 is negative, this happens when the bracket = 0, so when x=205.


EDIT

Or could differentiate it by the chain rule and solve it equal to 0 (turning point), but that isn't necessary.
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+1
Last edited by gm jack at Mar 1, 2008,
#6
uh.....im not understanding
killed off the y to get it equal to 0 and find out hte x-intercepts
but once its
-119 = -6/289(x-205)^2

i dont know what to do
Quote by WickedBeast666
Noooooooooooo how could this be! he at all the chocolate in the box
Oh well, now the empty shell of what used to be chocolaty goodness can contain a tasty guitar circuit.


speaking of my homemade pedal
#8
gm jack is right. (x-205)^2 >= 0 always. As such, when you multiply it by -6/289 it will always be <=0. Hence the maximum value of y (its highest peak) is when -6/289(x-205)^2 is minimised i.e. when -6/289(x-205)^2=0, so when x=205.
#9
k, if you solve through it simplifies to -

-6/289x^2 + 2460/289x + 991.49

so dy
dx =

-12/289x + 2460/289

so when the gradient = 0

the point is at 2460/289

i think, someone whould have to check that methodology
#11
Quote by fosho
uh.....im not understanding
killed off the y to get it equal to 0 and find out hte x-intercepts
but once its
-119 = -6/289(x-205)^2

i dont know what to do


Quote by muse_
I got x = 205, y = 3491 by completing the square.


how come your trying to find the intercepts ? the highest point could be anywhere on x

that type of curve you want to find where the gradient = 0 as thats where your top is
#12
This equation is already in vertex form, not standard form. The vertex is at (205, 119). Therefore the greatest y value is 119 and it occurs at x=205.
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#13
First things first. Develop this equation to the usual format of a second degree equation.

Then use the formulas to find the coordinates of the maximum point. ( -b/2a, for the maximum X, and -delta/4a for the maximum Y )

delta = b^2 - 4ac
#14
im a 10th grader in an 8th grade math class (seriously) so if i helped you, you'd get an F


for the record, not only am i in an 8th grade math class, i also have a C in said class....and i actually try too
UG's HIPPIE
#15
Quote by Def
how come your trying to find the intercepts ? the highest point could be anywhere on x

that type of curve you want to find where the gradient = 0 as thats where your top is

I wasn't trying to find the intercept. I didn't complete it all the way but got it into a form and read it off. I think I've done some dodgey calculating though cause I always **** up on the computer calculator.
#16
Write as the answer, "Is this question really real? Can there ever be a solution? Discuss". A's all the way for that.
#17
graph that ho?
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#18
its saturday night an your doin maths homweoKrk?

i shouldt omplaing though, yeserterday night i was in the darkroom
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#21
Look, if people cannot see how that is done, they need to reread my post.

It is based on the fact that at your level, and square number is greater tan or equal to 0.

Factor in the minus sign on the squared part and it means that number has to be less than or equal to 0.

We are trying to maximise y. Therefore, we don't let it become negative, and keep it at 0 by using x=205. The peak is y=119, as that is all that is left once you put x=205 in.

It is an inverted bucket shape. Any value of x greater or less than 205 will give a smaller value. Try it out is you are too stupid to see the logic in the working.


And this ain't calculus. Sure you can differentiate (need chain rule, but what people have said before is correct) to find the x=205 as the truning point, then put it in, but this is completing the square. Far simpler.
Warwick freak of the Bass Militia. PM Nutter_101 to join

Quote by elliott FTW
Damn you and Warwickyness

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#22
Oh ****. I've got rid of that fraction, completed the square and read it off whilst doing some calcs wrong as well. So basically done a load of **** to achieve what was already there 289x too big. Woops.

^Yeah that's right. Sorry.