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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 \(f\)?

 Feb 11, 2018
 #1
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f(-1)  = 15

f(0)  = 0

f(1)  =  -5

f(2)  = 12

 

A cubic  has the form  ax^3  + bx^2 + cx + d

Since (0,0)  is on the graph, then d  must equal 0

 

So.....we have this system

 

-a + b - c  =  15

a + b + c  =  -5

8a + 4b + 2c  = 12

 

-a + b - c  =  15

a + b + c    = -5      add these and we get that   2b = 10  ⇒ b = 5

 

And

 

8a + 5(4) + 2c  = 12

8a + 2c  =  -8   using this  and   -a + b - c  = 15   we have that

 

8a + 2c  =  -8

-a + 5 - c  = 15

 

8a + 2c  =  -8    (1)

-a - c   = 10       ⇒  -2a - 2c = 20  (2)     add (1)  and (2)

 

6a  =  12      ⇒  a  =  2

 

And using   - a + b - c  = 15

 

2 + 5 + c  = -5 ⇒   c = - 12

 

So.... the function is  

 

2x^3  + 5x^2  - 12x   .......to find the x intercepts.....set this to 0

 

2x^3  + 5x^2 - 12x  =  0      factor

 

x (2x^2 + 5x - 12)  = 0

 

x (2x - 3) (x + 4)  = 0

 

Setting the factors to 0   and solving for x we get that the x intercepts are  

 

 x = -4 , x  = 0   and x  =  3/2   = 1.5

 

 

cool cool cool

 Feb 11, 2018

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