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The polynomial  f(x) has degree 3. If f(-1)=15, f(0)=0, f(1)=-5, and f(2)=12, then what are the x-intercepts of the graph of f(x)?

 Mar 25, 2019
 #1
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We have the form

 

y = ax^3  + bx^2   + cx  +  d

 

Because f(0)  = 0.....then d  =  0

 

And we have this system

a(-1)^3 + b(-1)^2 + c(-1) = 15

a(1)^3 + b(1)^2 + c(1)   = 5

a(2)^3 + b(2)^2 + c(2)  = 12

 

 

-a + b - c  =15  (1)

a  + b + c   = 5   (2)  

8a   + 4b + 2c  = 12     (3)

 

Add the first two equations  and we get that   2b = 20  ⇒  b = 10

 

Multiply (2)  by -2  ⇒  -2a -2b -2c = -10

Add this to (3)

 

6a + 2b = 2

6a + 2(10) = 2

6a = -18

a = -3

 

And using   (2)

 

-3 + 10 + c = 5

c = -2

 

So....the polynomial is

 

f(x)  =  -3x^3 + 10x^2  - 2x  

 

This graph shows the x intercepts :  https://www.desmos.com/calculator/0qrtw2nob8

 

 

cool cool cool

 Mar 25, 2019

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