<h3>
Answer: n = -11</h3>
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Explanation:
Since x-2 is a factor of f(x), this means f(2) = 0.
More generally, if x-k is a factor of p(x), then p(k) = 0. This is a special case of the remainder theorem.
So if we plugged x = 2 into f(x), we'd get
f(x) = x^3+x^2+nx+10
f(2) = 2^3+2^2+n(2)+10
f(2) = 8+4+2n+10
f(2) = 2n+22
Set this equal to 0 and solve for n
2n+22 = 0
2n = -22
n = -22/2
n = -11 is the answer
Therefore, x-2 is a factor of f(x) = x^3+x^2-11x+10
Plug x = 2 into that updated f(x) function to find....
f(x) = x^3+x^2-11x+10
f(2) = 2^3+2^2-11(2)+10
f(2) = 8+4-22+10
f(2) = 0
Which confirms our answer.
The slope formula is y2-y1/x2-x1 so 5-(-1) =6 Then. 1-(-1)=2 so 6/2. =3. THE ANSWER IS 3
The period will be the length for one cycle of the function.
answer: 10
