Answer :
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:
[tex](x, y)=(\frac{-b}{2a}, \frac{-\Delta}{4a})[/tex]
a is 1, b is 6, c is 4, [tex]\Delta=b^2-4ac=36-16=20[/tex], 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.
[tex](x, y)=(\frac{-b}{2a}, \frac{-\Delta}{4a})[/tex]
a is 1, b is 6, c is 4, [tex]\Delta=b^2-4ac=36-16=20[/tex], 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)
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)
