Determine whether there is a maximum or minimum vale for the given function, and find that value. f(x)= x^2+6x+4

Every second degree function has either a maximum (if a is negative) or a minimum (if a is positive). Our function, thus, has a minimum. The formula for it is:
\( (x, y)=(\frac{-b}{2a}, \frac{-\Delta}{4a}) \)
a is 1, b is 6, c is 4, \( \Delta=b^2-4ac=36-16=20 \), so the minimum is at coordinates (-3,5), that is the function doesn’t ever get below -5 and it gets there only when the argument is -3.

If you take the first derivative of f(x) you get:

f’(x) = 2x + 6

The max or min is where the derivative is 0.

So. 0 = 2x + 6

Do some algebra and x = -3

Substitute that back into the original equation:

f(-3) = (-3)^2 + 6(-3) + 4

And you get -5

So your value is at the point (-3,5)