Answer:
Cubic polynomial has zeros at x=−1x=−1 and 22, is tangent to x−x−axis at x=−1x=−1, and passes through the point (0,−6)(0,−6).
So cubic polynomial has double zero at x=−1x=−1, and single zero at x=2x=2
f(x)=a(x+1)2(x−2)f(x)=a(x+1)2(x−2)
f(0)=−6f(0)=−6
a(1)(−2)=−6a(1)(−2)=−6
a=3a=3
f(x)=3(x+1)2(x−2)f(x)=3(x+1)2(x−2)
f(x)=3x3−9x−6
Answer:
84 degrees
Step-by-step explanation:
132+54 = 186 degrees
186+90 = 276 degrees
as this is around a point it totals to 360 degrees
360 degrees - 276 degrees = 84 degrees
hope this helps
I can't show the work, but if the problem is printed or on the computer, I recommend using Photomath in the AppStore :)
9514 1404 393
Answer:
Step-by-step explanation:
The extrema will be at the ends of the interval or at a critical point within the interval.
The derivative of the function is ...
f'(x) = 3x² -4x -4 = (x -2)(3x +2)
It is zero at x=-2/3 and at x=2. Only the latter critical point is in the interval. Since the leading coefficient of this cubic is positive, the right-most critical point is a local minimum. The coordinates of interest in this interval are ...
f(0) = 2
f(2) = ((2 -2)(2) -4)(2) +2 = -8 +2 = -6
f(3) = ((3 -2)(3) -4)(3) +2 = -3 +2 = -1
The absolute maximum on the interval is f(0) = 2.
The absolute minimum on the interval is f(2) = -6.