Y varies partly as x and partly as the square of x when x =2 and y=14 and when x=5 y=65, express y in terms of x hence find y when x=2
y = mx^2 + b
using the given data, you have
4m+b = 14
25m+b = 65
Now solve as usual for m and b
Not sure why you want to find y when x=2, since you already told us it was 14.
To express y in terms of x, we can assume that y varies partly as x and partly as the square of x.
Let's set up the equation:
y = kx + m(x^2)
where k and m are constants.
We can solve for k and m using the given information:
When x = 2 and y = 14:
14 = k(2) + m(2^2)
14 = 2k + 4m ----(1)
When x = 5 and y = 65:
65 = k(5) + m(5^2)
65 = 5k + 25m ----(2)
Now, we can solve the system of equations (1) and (2) to find the values of k and m.
Subtract equation (2) from equation (1):
14 - 65 = 2k - 5k + 4m - 25m
-51 = -3k - 21m
Divide by -3:
17 = k + 7m ----(3)
Now, we have an equation (3) with k and m. We can substitute this equation into (1) to find the value of k.
14 = 2k + 4m
Substitute k = 17 - 7m:
14 = 2(17 - 7m) + 4m
14 = 34 - 14m + 4m
14 = 34 - 10m
Isolate m:
-10m = -20
m = 2
Substitute m = 2 into equation (3) to find k:
17 = k + 7(2)
17 = k + 14
k = 17 - 14
k = 3
Now we have the values of k = 3 and m = 2.
So the expression for y in terms of x is:
y = 3x + 2(x^2)
To find y when x = 2, substitute x = 2 into the expression:
y = 3(2) + 2(2^2)
y = 6 + 2(4)
y = 6 + 8
y = 14
Therefore, when x = 2, y = 14.