Answer:
The vertex form is f(x) = (x + 5)² + 12
The minimum value of f(x) is point (-5, 12)
Explanation:
1) The vertex form of a quadratic equation f(x) = x² + 10·x + 37, which is the form f(x) = a·(x - h)² + k is found as follows;
For the general form of the quadratic equation, f(x) = a·x² + b·x + c
h = -b/(2·a) and k = f(h)
Therefore, for f(x) = x² + 10·x + 37,
a = 1, b = 10
∴ h = -10/2 = -5
k = f(-5) = (-5)² + 10×(-5) + 37 = 12
The vertex form is f(x) = a·(x - h)² + k gives;
f(x) = 1·(x - (-5))² + 12 = (x + 5)² + 12
The vertex form is f(x) = (x + 5)² + 12
2) The minimum value of x is found when d(f(x))/dx = 0
d(f(x))/dx = d(x² + 10·x + 37)/dx = 2·x + 10
d(f(x))/dx = 0 = 2·x + 10
x = -10/2 = -5
We check that it is the minimum by f''(x) being positive;
f''(x) = d(2·x + 10)/dx = 2 which is positive and x = -5 is the x-coordinate of the minimum value of f(x)
The x-coordinate of the minimum value of f(x) minimum value is f(-5) = (-5)² + 10×(-5) + 37 = 12
Therefore, we have;
The minimum value of f(x) = (-5, 12)