Graph the function by first finding the relative extrema.
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f(x) = x^3 + 4x^2 - x - 4
f'(x) = 3x ^2 + 8x -1
c= 3x ^2 + 8x -1
Using the quadratic equation
[tex]x=\frac{-b\text{ }\pm\text{ }^{}\sqrt[]{b^2\text{ -4ac}}}{2a}\text{ = }\frac{-(8)\text{ }\pm\text{ }^{}\sqrt[]{8^2\text{ -4}\cdot3\cdot\text{ (-1)}}}{2\cdot3}[/tex]
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They want you to see the extreme points, but the easiest way is to evaluate 0 and check which graph matches
f(0) = 0^3 + 40^2 - 0 - 4
Point (0, -4)
