How do you sketch the curve #y=x^3+6x^2+9x# by finding local maximum, minimum, inflection points, asymptotes, and intercepts?
X 2 5 y 5 3 2 2 Answers ar 6cos θ r 2 6 rcos θ x2 y 2 6x x2 6 xy 2 x 2 6 x 9 y from MATH 004062 at Pioneer High School. Study Resources. Main Menu; by School; by Textbook; by Literature Title. Study Guides Infographics. Expert Tutors Contributing. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history.
1 Answer
Explanation:
Split screen 3 4 download free. Polynomial graphs have no asymptotes.
As x to +-oo, y = x^3(1+3/x)^2 to +-oo, showing end behavior of Imusic 2 2 0 4.
So, there are no global extrema.
x-intercepts: #0 and -3# .
Turning points or points of inflexion at (-1, -4) and (-3, 0)
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At #(-2, -2), y'=0 and y'' ne 0# . So, it is the point of inflexion.
This POI is marked, in the graph.
At # (-1, -4)), y'=0 and y'=6> 0# . So, local minimum #y = -4# .
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At #(-3, 0), y'=0 and y'= -6 # . So,#0# is a local maximum.
graph{(x(x+3)^2-y)((x+2)^2+(y+2)^2-.01)=0 [-10, 10, -10, 5]}
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