y = kx^2 + c
when x=1, y=11 ----> 11 = k + c
when x=2, y=5 ----> 5 = 4k + c
subtract them:
3k = -6
k = -2
in k+c=11
-2+c=11
c = 13
so y = -2x^2 + 13
now simply find y when x = 4
The quantity Y is partly constant and partly varies as the square of X (a) write down the relationship between X and Y. (b) when X is 1,Y is11 and when X is 2 and Y is 5 find the value of Y when X is 4.
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