Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
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)
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
