Answer:
[tex] f(x) = 4x^2 - 3x + 6 [/tex]
Step-by-step explanation:
Quadratic function is given as [tex] f(x) = ax^2 + bx + c [/tex]
Let's find a, b and c:
Substituting (0, 6):
[tex] 6 = a(0)^2 + b(0) + c [/tex]
[tex] 6 = 0 + 0 + c [/tex]
[tex] c = 6 [/tex]
Now that we know the value of c, let's derive 2 system of equations we would use to solve for a and b simultaneously as follows.
Substituting (2, 16), and c = 6
[tex] f(x) = ax^2 + bx + c [/tex]
[tex] 16 = a(2)^2 + b(2) + 6 [/tex]
[tex] 16 = 4a + 2b + 6 [/tex]
[tex] 16 - 6 = 4a + 2b + 6 - 6 [/tex]
[tex] 10 = 4a + 2b [/tex]
[tex] 10 = 2(2a + b) [/tex]
[tex] \frac{10}{2} = \frac{2(2a + b)}{2} [/tex]
[tex] 5 = 2a + b [/tex]
[tex] 2a + b = 5 [/tex] => (Equation 1)
Substituting (3, 33), and c = 6
[tex] f(x) = ax^2 + bx + x [/tex]
[tex] 33 = a(3)^2 + b(3) + 6 [/tex]
[tex] 33 = 9a + 3b + 6 [/tex]
[tex] 33 - 6 = 9a + 3b + 6 - 6 [/tex]
[tex] 27 = 9a + 3b [/tex]
[tex] 27 = 3(3a + b) [/tex]
[tex] \frac{27}{3} = \frac{3(3a + b)}{3} [/tex]
[tex] 9 = 3a + b [/tex]
[tex] 3a + b = 9 [/tex] => (Equation 2)
Subtract equation 1 from equation 2 to solve simultaneously for a and b.
[tex] 3a + b = 9 [/tex]
[tex] 2a + b = 5 [/tex]
[tex] a = 4 [/tex]
Replace a with 4 in equation 2.
[tex] 2a + b = 5 [/tex]
[tex] 2(4) + b = 5 [/tex]
[tex] 8 + b = 5 [/tex]
[tex] 8 + b - 8 = 5 - 8 [/tex]
[tex] b = -3 [/tex]
The quadratic function that represents the given 3 points would be as follows:
[tex] f(x) = ax^2 + bx + c [/tex]
[tex] f(x) = (4)x^2 + (-3)x + 6 [/tex]
[tex] f(x) = 4x^2 - 3x + 6 [/tex]
