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.
+128408 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
