is there a value to y? if not, this is what I got:
Y^2 + 2y + 3
Yes is the answer click it
60
2 and 30
30
2 and 15
15
3 and 5
96
8 and 12
circle for 8 is 2 and the square is 4
circle for 12 is 2 and the last circle is also 2
Step-by-step explanation:
Take the first derivative
Set the derivative equal to 0.
or
For any number less than -1, the derivative function will have a Positve number thus a Positve slope for f(x).
For any number, between -1 and 1, the derivative slope will have a negative , thus a negative slope.
Since we are going to Positve to negative slope, we have a local max at x=-1
Plug in -1 for x into the original function
So the local max is 2 and occurs at x=-1,
For any number greater than 1, we have a Positve number for the derivative function we have a Positve slope.
Since we are going to decreasing to increasing, we have minimum at x=1,
Plug in 1 for x into original function
So the local min occurs at -2, at x=1
Point form: (-2,-3)
Equation form: x=-2, y=-3