anonymous
  • anonymous
I need someone to help me (in very simple terms ;)) how to: Graph each polynomial function. Factor first if the expression is not in factored form.. f(x)=x^2(x+1)(x-1)
Mathematics
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anonymous
  • anonymous
I need someone to help me (in very simple terms ;)) how to: Graph each polynomial function. Factor first if the expression is not in factored form.. f(x)=x^2(x+1)(x-1)
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
to graph polynomial function , firstly i will find the point which intersect x-axis. the point that intersect x-axis has y=0 to find that, 0=x^2 (x+1)(x-1) the values of x which fulfill the equation are x=0, x=-1, x=1 secondy, find the extreme point. use the derivative of the function \[f \prime x =4x ^{3}-2x ^{}=0\] so x=-1/2 and 1/2 then plug those value of x to the function to get the extreme value of y you'll get y = 1/16 so you'll have some points : (0,0) , (-1,0) , (1,0) (-1/2, 1/16), (1/2, 1/16) finally just connect that points so now you have the graph
anonymous
  • anonymous
sorry, the value of y extreme is not 1/16 . it's -1/16
anonymous
  • anonymous
so the extreme points is (-1/2, -1/16) and (1/2, -1/16)

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anonymous
  • anonymous
here, i attach the graph of the function
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anonymous
  • anonymous
Thank you so much!!!!!!!!!! Great, great explanation!
anonymous
  • anonymous
you're welcome :)

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