Discover new information and insights with the help of IDNLearn.com. Join our interactive community and access reliable, detailed answers from experienced professionals across a variety of topics.
Sagot :
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)
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
