Let f(x)=x2+10x+37 . What is the vertex form off(x)? What is the minimum value off(x)? Enter your answers in the boxes. Vertex form: f(x)= Minimum value of f(x):

The vertex form is f(x) = (x + 5)² + 12

The minimum value of f(x) is point (-5, 12)

Step-by-step 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)

1/4 because there is 4 grades.