(a).Time (t) 0 weeks 3 weeks 6weeks
Investment Value v(t) $10000 $2025 $6400
Using the data written down in the table in (a) derive a quadratic function (you BiQuad model), 𝑣(𝑡), relating the value of Peter’s share portfolio to elapsed time. Show all working.
Not sure what BiQuad is, but if you want to fit a parabola through those three points, you have, assuming that
v(t) = at^2 + bt + c,
0a+0b+c = 10000
9a+3b+c = 2025
36a+6b+c = 6400
Somehow I suspect what you really wanted was v(3) = 1000-2025 = 7975, since most stocks don't bounce like that.
At any rate, solve for a,b,c for your model:
clearly, c=10000
Now just find a and b.
To derive a quadratic function relating the value of Peter's share portfolio to elapsed time, we can use the given data points from the table and solve for the quadratic equation.
Let's consider the general quadratic function:
v(t) = at^2 + bt + c
We need to find the values of coefficients a, b, and c using the given data points.
From the table, we have the following data:
t = 0 weeks, v(t) = $10000
t = 3 weeks, v(t) = $2025
t = 6 weeks, v(t) = $6400
Substituting these values into the quadratic function, we get three equations:
1st equation: 10000 = a(0)^2 + b(0) + c
10000 = c -- (i)
2nd equation: 2025 = a(3)^2 + b(3) + c
2025 = 9a + 3b + c -- (ii)
3rd equation: 6400 = a(6)^2 + b(6) + c
6400 = 36a + 6b + c -- (iii)
Now, we have a system of three equations (i), (ii), and (iii) that we can solve simultaneously to find the values of a, b, and c.
Subtracting equation (ii) from equation (i), we get:
7975 = 9a + 3b -- (iv)
Subtracting equation (iii) from equation (i), we get:
3600 = 36a + 6b -- (v)
We now have a system of two linear equations (iv) and (v) that we can solve to find the values of a and b.
Solving equations (iv) and (v), we get:
a = -25/27 and b = 325/9
Finally, substituting these values of a and b in equation (i) or (ii) will give us the value of c:
10000 = c
So, the quadratic function relating the value of Peter's share portfolio to elapsed time is:
v(t) = (-25/27)t^2 + (325/9)t + 10000