+0  
 
0
80
1
avatar+2720 

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 \(f\)?

tertre  Feb 11, 2018
 #1
avatar+86939 
+1

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

CPhill  Feb 11, 2018

8 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.