Answer:
The standard form of the quadratic equation is f(x) = 2/3·(x + 2)² + 6
Explanation:
The given solution of the quadratic equation is x = -2 + 3·i
A point on the path of the quadratic equation = (-1, -15)
The general form of the quadratic equation is f(x) = a·x² + b·x + c
When f(x) = 0, (At which the solution is found), we have x = -2 + 3·i
Substituting gives;
0 = a·(-2 + 3·i)² + b·(-2 + 3·i) + c
0 = 4·a - 12·a·i - 9·a - 2·b + 3·b·i + c
0 = 3·i·(b - 4·a) - 5·a - 2·b + c
∴ b = 4·a
c = 13·a
The general equation becomes
f(x) = a·x² + 4·a·x + 13·a
Also, when x = -1, f(x) = -15
f(-1) = -15 = a·(-1)² + 4·a·(-1) + 13·(-1)
-15 = a - 4·a - 13 = -3·a - 13
-3·a = -15 + 13 = -2
a = 2/3
The general form of the quadratic equation is therefore;
f(x) = 2/3·x² + 8/3·x + 13·2/3 = 2/3·x² + 8/3·x + 26/3
f(x) = 2/3·x² + 8/3·x + 26/3
The minimum value occurs at d(f(x)/dx = 0 = d(2/3·x² + 8/3·x + 13·2/3)/dx = 4/3·x + 8/3 = 0
x = -8/3 × 3/4 = -2
Therefore;
The minimum value of the quadratic function is f(-2) = 2/3·(-2)² + 8/3·(-2) + 13·2/3 = 6
The coordinates of the minimum point, which is the vertex point (h, k) = (-2, 6)
The standard form of the quadratic equation f(x) = a(x - h)² + k, is therefore;
f(x) = 2/3(x - (-2))² + 6 = 2/3·(x + 2)² + 6
f(x) = 2/3·(x + 2)² + 6
Also (h, k) can be gotten from h = -b/2a = -8/3/(2 × 2/3) = -2
k = c - b²/(4·a) = 26/3 - (8/3)²/(4 × 2/3) = 26/3 - 8/3 = 18/3 = 6
k = 6
The standard form of the quadratic equation f(x) = a(x - h)² + k becomes;
f(x) = 2/3·(x + 2)² + 6