y is party constant and partly varies
as square of x when x =1, y=11 and when x=2, y=5, find the value of y when x=4.
y = a x^2 + b
11 = a + b
5 = 4 a + b
-----------------
6 = -3 a
a = -2
b = 13 etc
To find the value of y when x=4, we will use the given information that y is constant and partly varies as the square of x.
Given that when x=1, y=11 and when x=2, y=5.
Let's denote the constant part of y as "c" and the variable part as "k". Therefore, we can represent y as:
y = c + k*x^2
Substituting the values x=1 and y=11 into the equation:
11 = c + k*1^2 -> 11 = c + k
Substituting the values x=2 and y=5 into the equation:
5 = c + k*2^2 -> 5 = c + 4k
Now we have two equations:
Equation 1: 11 = c + k
Equation 2: 5 = c + 4k
By solving these two equations simultaneously, we can find the values of c and k.
Subtracting Equation 2 from Equation 1:
11 - 5 = c + k - (c + 4k)
6 = k - 3k
6 = -2k
k = -3
Replacing the value of k in Equation 2:
5 = c - 12
By solving for c:
c = 5 + 12 = 17
Now we have the values of c and k:
c = 17
k = -3
Now, we can find the value of y when x=4 by substituting these values into the equation:
y = c + k*x^2
y = 17 + (-3)*4^2
y = 17 + (-3)*16
y = 17 - 48
y = -31
Therefore, when x=4, the value of y is -31.
To find the value of y when x=4, we need to determine the relationship between x and y based on the given information.
From the problem, we are told that y is constant and partly varies as the square of x. This means that y can be expressed as the sum of a constant term and a term that involves x squared.
Let's denote the constant term as c and the term involving x squared as kx^2.
We can then write the equation for y as:
y = c + kx^2
Now, we are given two specific points: When x = 1, y = 11, and when x = 2, y = 5.
Using these points, we can create a system of equations to solve for the values of c and k.
When x = 1, y = 11:
11 = c + k(1^2) = c + k
When x = 2, y = 5:
5 = c + k(2^2) = c + 4k
Now, we can solve this system of equations to find the values of c and k.
Firstly, subtract the first equation from the second equation to eliminate c:
5 - 11 = (c + 4k) - (c + k)
-6 = 3k
k = -2
Substitute the value of k back into the first equation:
11 = c + (-2)
11 = c - 2
c = 13
Now that we have found the values of c and k, we can plug them back into the equation for y:
y = 13 - 2x^2
Finally, plug in x = 4 to find the value of y:
y = 13 - 2(4^2)
y = 13 - 2(16)
y = 13 - 32
y = -19
Therefore, when x = 4, the value of y is -19.