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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?

 Mar 29, 2019
 #1
avatar+106080 
+2

I have not worked through this qustion but I'd start with the generic polynomial

 

\(f(x)=ax^3+bx^2+cx+d\)

 

Then start substituting so see if you can get any of those coefficients.

Starrt with  f(0)=0

 

You will probably end up with a couple of equations that you will need to solve simultaneously.

 

See how you go and post what you find for yourself.

If you then need more help then continue this thread.

 

Please no one else answer until asker guest has returned with some input of his/her own.

 Mar 29, 2019
edited by Melody  Mar 29, 2019
 #2
avatar+16 
+2

It has no constant because f(0)=0

c_3x^3+c_2x^2+c_1x

-c_3+c_2-c_1=15

c_3+c_2+c_1=-5

8c_3+4c_2+2c_1=12

4c_3+2c_2+c_1=6

-2c_3+2c_2-2c_1=30

*2

6c_3+6c_2+6c_1=-30

*6

add together

4c_3+8c_2+4c_1=0

divide by 4

c_3+2c_2+c_1=0

compare for x intercepts

c_3x^3+c_2x^2+c_1x=c_3+2c_2+c_1

 Mar 30, 2019
 #3
avatar+106080 
+1

That is rather difficult to decipher but have you worked it out?

Melody  Mar 30, 2019
 #4
avatar+16 
+2

I've figured out the answer to the question. I just had to do some substitution with the equations. The x-intercepts are (-4,0), (0,0) and (1.5,0). =)

 Mar 31, 2019
 #5
avatar+106080 
+2

So what is the polynomial?  Did you graph it using desmos to check your answer?

 

https://www.desmos.com/calculator

Melody  Mar 31, 2019
 #6
avatar+16 
+1

Yep. Here is the link to the graph: https://www.desmos.com/calculator/bvpzoqrgnu

 Mar 31, 2019
 #7
avatar+106080 
+1

GOOD WORK !!    laugh

 

I added to your graph :)

 

https://www.desmos.com/calculator/astkapa4gb

Melody  Mar 31, 2019

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