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.
(a) To represent the relationship between X and Y, we can write it as:
Y = kX + cX^2
where k represents the constant part and c represents the coefficient of X^2 that reflects the variation.
(b) Given that when X is 1, Y is 11 and when X is 2, Y is 5, we can use these values to find the values of k and c in the equation.
When X is 1,
11 = k(1) + c(1)^2
11 = k + c
When X is 2,
5 = k(2) + c(2)^2
5 = 2k + 4c
We have a system of equations that can be solved simultaneously to find the values of k and c. Subtracting the first equation from the second, we get:
5 - 11 = 2k + 4c - (k + c)
-6 = k + 3c
Substituting the value of k from the first equation into this new equation:
-6 = (11 - c) + 3c
-6 = 11 - 2c
2c - 6 = 11
2c = 17
c = 8.5
Substituting the value of c back into the first equation to find k:
11 = k + (8.5)
11 = k + 8.5
k = 11 - 8.5
k = 2.5
Now that we have the values of k and c, we can use them in the original equation to find Y when X is 4:
Y = (2.5)(4) + (8.5)(4^2)
Y = 10 + 8.5(16)
Y = 10 + 136
Y = 146
Therefore, when X is 4, Y is 146.
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