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The polynomial has degree 3. If f(-1) = 32, f(0) = 0, f(1) = -24, ,and f(2) = -28, then what are the x-intercepts of the graph of f?

 Nov 14, 2020
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We have the form  P(x) = ax^3 + bx^2 + cx^2  + d

 

Since f(0) =  0  .....then d   =  0

 

So we actually  have this system of equations

 

a(-1)^3  + b(-1)^2  + c(-1) = 32

a(1)^3  + b(1)^2   +  c(1)  = -24

a(2)^3  + b(2)^2  + c(2)  = -28           simplify these

 

-a + b  - c  =  32   (1)

 a  + b + c  = -24   (2)

8a  + 4b  + 2c  =  -28    (3)

 

Add (1)  and (2)

2b = 8

b = 4

 

Sub this into  (2) and (3)

a + 4 + c  =-24 →  a + c  = -28   (4)

8a + 4(4) + 2c  = -28  →   8a + 2c = -44→  4a + c = -22  →  -4a - c = 22   (5)

 

Add (4)  and (5)

-3a = -6

a = 2

 

Using (4)

2 + c  = -28

c  = -30

 

Our polynomial is

 

P(x)  =  2x^3 + 4x^2 - 30x

 

Setting this to 0  and dividing through by 2, we have that

 

x^3 + 2x^2  -15x  = 0           factor

 

x (x^2 + 2x - 15)  =  0

 

x ( x -3) (x + 5)  = 0

 

Setting  each factor to 0  and  solving for x, the intercepts are

x = 0    x = 3  and  x  = -5

 

See the graph here : https://www.desmos.com/calculator/zrdafdxpea

 

 

cool cool cool

 Nov 14, 2020

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