Answer:
An example of the quadratic function that has zeros at x = 2 and x = 4, and that takes the value 6 when x = 3 is f(x) = -6·x² + 36·x 48
Step-by-step explanation:
The general form of a quadratic function is f(x) = a·x² + b·x + c
Which gives;
When x = 3, f(x) = 6 which gives;
6 = a·3² + b·3 + c = a·9 + b·3 + c............(1)
The zeros of the function are x = 2 and x = 4 which gives
0 = a·2² + b·2 + c = a·4 + b·2 + c...............(2)
0 = a·4² + b·4 + c = a·16 + b·4 + c...............(3)
Subtracting equation (2) from (3) gives;
a·16 + b·4 + c - (a·4 + b·2 + c) = a·12 + b·2 = 0
b·2 = -a·12
b = -a·12/2 = -6·a
Substituting b = -6·a in equation (1) we have;
a·9 + b·3 + c = a·9 + (-6·a)·3 + c = 6
a·9 -18·a + c = 6
-9·a + c = 6
c = 6 + 9·a
Substituting b = -6·a and c = 6 + 9·a in equation (2), we get
0 = a·4 + b·2 + c = a·4 + (-6·a)·2 + 6 + 9·a = a·4 -12·a + 6 + 9·a
a + 6 = 0
a = - 6
b = -6·a = -6*(-6) = 36
c = 6 + 9·a = 6 + 9*(-6) = 6 - 54 = -48
The equation of the function is therefore, f(x) = -6·x² + 36·x 48.
