First thing is to check on the wording of the question.
The two areas are the same aren't they ?
I have a curve, x^3 - 14x^2 + 59x -70 , and I need to run this line through it, mx - 70 , to make the area above the curve enclosed by the line equal to HALF of the area below the line enclosed by the curve..... So, I need to find the gradient (m) of the line to run through it for this to be possible.
So far I have derived that the x intersects are :
x=0
x= 7 - square root of m-10
x= 7 + square root of m-10
WHAT NOW
So the area above the line and below the curve is and the area below the line and above the curve is . You need to calculate these two integrals (treating m as a constant), then set the second equal to half of the first and solve for m.
The process is straightforward, but kind of messy.
- Hollywood
The x-intercepts of what? There are two graphs refered to, the graph of y= x^3- 14x^2+ 59x- 70 and y= mx- 70 and neither has what you give as x- intercepts. Do you mean where the graphs intersect, not "x-interepts"?
x=0
x= 7 - square root of m-10
x= 7 + square root of m-10
WHAT NOW
Here's what you should get:
so setting the second to half of the first gives
and the solution is
- Hollywood
P.S. jomex2: You should start a new thread with your problem. Few (if any) people besides Bryanna and me are reading this thread now.
ok so I am starting to see where you are coming from... But... you know how you did the first bit = - m^2/4 + 14m/3 (square root) m-10..... That answer, how did you get it? I am expanding it and I dont get how you got that