Determine whether the quadratic function shown below has a minimum or maximum, then determine the minimum or maximum value of the function.
f(x)= (x+4)(x+8)
First, let's expand the function:
f(x) = (x+4)(x+8)
f(x) = x^2 + 8x + 4x + 32
f(x) = x^2 + 12x + 32
This is a quadratic function in the form f(x) = ax^2 + bx + c. Since the coefficient of x^2 is positive (a = 1), the parabola opens upwards and therefore has a minimum value.
To find the minimum value of the function, we can use the vertex formula:
The x-coordinate of the vertex is given by:
x = -b / (2a)
x = -12 / (2*1)
x = -12 / 2
x = -6
Now, substitute x = -6 back into the function to find the minimum value:
f(-6) = (-6)^2 + 12*(-6) + 32
f(-6) = 36 - 72 + 32
f(-6) = -4
Therefore, the minimum value of the function f(x) = x^2 + 12x + 32 is -4, and it occurs at x = -6.