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

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

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