SOLUTION: Hi, I'm having trouble with this question from my A level Integration module; Find the equation for a quadratic curve that passes through O(0,0) and its gradient at the poin

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Question 924934: Hi,
I'm having trouble with this question from my A level Integration module;
Find the equation for a quadratic curve that passes through O(0,0) and its gradient at the point (-2,3) is -1.
I can integrate a differentiated function to get it's original, but without the differential I'm stuck on this and can't see where to begin.
Thanks,
R
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Quadratic curve : y=ax%5E2%2Bbx%2Bc
Passes through the origin :
0=a%280%29%2Bb%280%29%2Bc
c=0
So,
y=ax%5E2%2Bbx
The derivative of the function is,
dy%2Fdx=2ax%2Bb
The function also passes through (-2,3) where,
dy%2Fdx=-1
2a%28-2%29%2Bb=-1
1.-4a%2Bb=-1
But also from the value of the function,
3=a%28-2%29%5E2%2Bb%28-2%29
2.4a-2b=3
Add eq. 1 to eq. 2,
-4a%2Bb%2B4a-2b=-1%2B3
-b=2
b=-2
Then,
-4a-2=-1
-4a=1
a=-1%2F4
So then,
highlight%28y=-%281%2F4%29x%5E2-2x%29