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.

Guest Oct 28, 2019

edited by
Guest
Oct 28, 2019

edited by Guest Oct 29, 2019

edited by Guest Oct 29, 2019

edited by Guest Oct 29, 2019

edited by Guest Oct 29, 2019

#1**+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

CPhill Oct 29, 2019