Explain how to sketch a graph of the function f(x) = x3 + 2x2 – 8x. Be sure to include end-behavior, zeroes, and intervals where the function is positive and negative.
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Answer:
x < - 4 f(x) is -ve
-4 < x < 0 f(x) is + ve
0 < x < 2 f(x) is -ve
x > 2 f(x) is +ve
Step-by-step explanation:
f(x) = x³ + 2x² - 8x
= x(x² + 2x -8)
= x( x² + 4x - 2x -8)
=x(x(x+4) -2(x+4))
=x(x-2)(x+4)
Zeroes are
0 , 2 , - 4
x < - 4 f(x) is -ve
-4 < x < 0 f(x) is + ve
0 < x < 2 f(x) is -ve
x > 2 f(x) is +ve
x f(x)
-5 -35
-4 0
-3 15
(-4-4√7)/6 16.9
-2 16
-1 9
0 0
1 -5
(-4+4√7)/6 -5.05
2 0
3 21
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