`y = x^3 - 6x^2 - 15x + 4` Sketch the curve by locating max/mins, asymptotes, points of inflection, etc.

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Chapter 4, Review - Problem 20 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`y=x^3-6x^2-15x+4`

a) Asymptotes

Polynomial function of degree 1 or higher can't have any asymptotes.

b) Maxima/Minima

`y'=3x^2-12x-15`

`y'=3(x^2-4x-5)`

`y'=3(x-5)(x+1)`

Now we can find critical numbers by solving x for y'=0,

`3(x-5)(x+1)=0rArrx=5 , x=-1`

Now let's check the sign of y' by plugging test points in the intervals (-`oo` ,-1) , (-1,5) and (5,`oo` )

`y'(-2)=3(-2-5)(-2+1)=21`

`y'(1)=3(1-5)(1+1)=-24`

`y'(6)=3(6-5)(6+1)=21`

Local Maximum at x=-1

`y(-1)=(-1)^3-6(-1)^2-15(-1)+4=12`

Local Minimum at x=5

`y(5)=5^3-6(5^2)-15(5)+4=-96`

c) Inflection Points

`y''=6x-12=6(x-2)`

Solve for x ,y''=0

`6(x-2)=0rArrx=2`

`y(2)=2^3-6(2^2)-15(2)+4=-42`

Inflection point is (2,-42)

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