Pls show work. Graph the curv e y = 1/3x3 - x2 - 3x + 5. Find the intercept. poi

Question

Pls show work.

Graph the curv e y = 1/3x3 - x2 - 3x + 5. Find the intercept. points where tangent is horizontal (critical point or point of extrema), where the graph is increasing and decreasing, where concave up and down, and inflection points. Label the preceding points on your graph.

Explanation / Answer

I can't graph it for you but I can do everything else. y-intercept: y = 5 The tangent is horizontal where the first derivative equals 0. So: y' = x^2 - 2x - 3 = (x-3)(x+1) => (x-3)(x+1) = 0 => x = 3 or x = -1 So the points where the tangent line is horizontal is at x = -1,3 y'(-2) = 4 + 4 - 3 > 0 y'(0) = -3 < 0 y'(1) = 1 - 2 - 3 < 0 So the graph is increasing @ (-infinity, -1) and it is decreasing @ (-1,infinity) The inflection points are where the second derivative equals 0: y'' = 2x - 2 => 0 = 2x - 2 => 2x = 2 => x = 1 So the inflection point is x = 1. y''(0) = -2 < 0 y''(2) = 4 - 2 > 0 So the graph is concave down @ (-infinity, 1) and concave up @ (1,infinity)

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