We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
179
1
avatar

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

 Jul 30, 2018
 #1
avatar+27667 
0

You can write the polynomial as:

 

f(x) = x(ax2+bx+c)     This form immediately gives f(0) = 0.

 

Now use the other conditions to find constants a, b and c:

 

15 = -1( (-1)2a + (-1)b +c)   or  15 = -a + b - c   ...(1)

 

-5 = 1(12a + 1b + c)  or  -5 = a + b + c    ...(2)

 

12 = 2(22a + 2b + c)  or  12 = 8a + 4b + 2c   or  6 = 4a + 2b + c...(3)

 

Add (1) and (2) to get:  10 = 2b  or  b = 5

 

Substitute this into (2) and (3) and rearrange:

-10 = a + c   ...(2b)

-4 = 4a + c  ...(3b)

 

Subtract (2b) from (3b)

6 = 3a   or  a = 2

 

Substitute this into (2b)

-10 = 2 + c  or  c = -12

 

Now we have:  f(x) = x(2x2 + 5x - 12) 

This factors as f(x) = x(2x - 3)(x + 4)

 

Hence the roots are: 0, 3/2, -4

 

You should check this by substituting the values -1, 0, 1 and 2 for x into the function to see they give the values specified in the question.

 Jul 31, 2018

22 Online Users

avatar