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
61
1
avatar

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?

 Aug 26, 2019
 #1
avatar+102913 
+2

Since f(0)  = 0,,,,then x  = 0  is one root

 

So.....we have the  form    ax^3 + bx^2 + cx + d   where d  = 0

 

So  we know that

 

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

a (1)^3  + b(1)^2 + c(1)  = -5   ⇒      a+ b + c = -5   (2)

a(2)^3 + b(2)^2 + c(2)  = 12  ⇒         8a + 4b  + 2c = 12    (3)

 

Add (1) and (2)  and we have that 2b = 10  ⇒   b  = 5

 

Multiply (1) by 8  and add to (3)   and we have that

 

12b - 6c  = 132     sub 5 in for b  and we have

12(5) - 6c  = 132

60 - 6c  = 132     subtract 60 from both sides

-6c  =  72   divider both sides by -6

c  = -12

 

And using (2)  we can find  a  as

 

a + 5  - 12  = -5

a - 7  =  -5    add 7 to both sides

a  = 2

 

So.....the polynomial  is     2x^3  + 5x^2 - 12x

 

Set to 0  and factor

 

x (2x^2 + 5x - 12)  = 0

 

x (2x  - 3) ( x + 4)  = 0      set each factor to 0   and solve for x

 

x  = 0         2x - 3  =  0        x + 4  = 0

 

x = 0       x  = 3/2          x  = -4       these  are the x intercepts

 

Here is a graph  : https://www.desmos.com/calculator/o040mctwqm

 

 

cool cool cool

 Aug 26, 2019

18 Online Users

avatar