# Find the cubic polynomial f(x)=ax^3+bx^2+cx+d that has horizontal tangents at the points (-1,-6) and (3,26).

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### 2 Answers

`f(x)= ax^3 + bx^2 + cx + d`

We need to fin a,b,c, and d such that f(x) has horizontal tangent lines at the point (-1,-6) and (3,26)

The slope of the horizental tangent line is 0.

=> f'(-1) = 0

==> f'(3) = 0

`==gt f'(x)= 3ax^2 + 2bx + c`

==> f'(-1) = 3a -2b + c = 0.........(1)

==> f'(3)= 27a +6b + c = 0 ........(2)

Subtract (1) from (2).

==> 24a + 8b = 0

Divide by 8.

==> 3a + b= 0...........(3)

Now we know that:

f(-1) = -6

`==gt f(-1)= a(-1)^3 + b(-1)^2 + c(-1) + d = -6`

==> -a + b -c + d = -6 ....................(4)

f(3)= 26

`==gt f(3)= a(3^3) + b(3^2)+c(3) + d = 26`

==> 27a + 9b +3c + d = 26 ............(5)

Now we will subtract (4) from (5)

==> 28a +8b + 4c = 32

Divide by 4.

==> 7a + 2b + c = 8 .............(6)

Now we will subtract (1) from (6)

==> 4a + 4b = 8

==> a + b= 2 ....................(7)

Now we will subtract (7) from (3)

`==gt 2a = -2`

`==gt a = -1 =`

`=gt b= 2-a = 2+1 = 3 `

`==gt c = 2b-3a = 2(3)- 3(-1) = 6+3 = 9 `

`==gt d = -6+c+a-b = -6+9 -1 - 3= -1 `

`==gt f(x)= -x^3 + 3x^2 + 9x -1`

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Find the cubic polynomial of f(x)=ax^3+bx^2+cx+d that has the horizontal tangents of (-3,3) & (4,-3).

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Ah, bollocks. Ignore that, meant to ask a question, heh.