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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?

CPhill, I see you're composing in answer but it has been taking pretty long so I just wanted to let you know I don't need the graph, I just need the x-intercepts.

Anyone else feel free to post an answer.

Oct 28, 2019
edited by Guest  Oct 28, 2019
edited by Guest  Oct 29, 2019
edited by Guest  Oct 29, 2019

#1
+109202
+2

We have the form

ax^3 + bx^2 + cx +d

I think that we should have  f(-1)  = 15

Since (0,0)  is on the graph, then d  =  0

So  we have this system of equations

-a   +  b  -  c   = 15    (1)

a  +  b  +   c  = -5      (2)

8a  + 4b  + 2c   = 12    ⇒   4a + 2b + c  =  6  (3)

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

Sub this into  (2) and (3) and we get that

a + 5 + c  =  -5  ⇒   a + c  =  -10   ⇒  -a - c  =  10   (4)

4a + 2(5)  + c  = 6  →  4a + c  =  -4     (5)

Add (4)  and (5)  and we get that

3a  = 6

a  = 2

And

-2 + 5 - c   = 15

-c  = 12

c  = -12

So....the function is

f(x)  = 2x^3  + 5x^2 - 12x

Factoring we have that

x (2x^2 + 5x - 12)  = 0

x (2x - 3)(x + 4)  = 0

So....settting  each factor to 0   and solving for x gives the  x intercepts  of  x  = 0 , x = 3/2 and x  = -4

Oct 29, 2019
#2
+1

thank you.

Guest Oct 29, 2019
#3
+2833
+1

You have been blessed by God Cphill :D:D:D:D

CalculatorUser  Oct 29, 2019
#4
+109202
+1

LOL!!!!!

CPhill  Oct 29, 2019
#5
0

Yes, but he was also cursed by the devil, so it mostly balances out.

Guest Oct 29, 2019
#6
+2833
+1

ahaha! "mostly"

CalculatorUser  Oct 29, 2019