The polynomial \(f(x)\) has degree \(3\). If \(f(-1)\) \(= 15\), \(f(0) = 0\) and \(f(1) = -5\) and \(f(2) = 12\) then what are the \(x\)-intercepts of the graph of \(f\)? Thank you!
+23245 The equation will be of this form: ax3 + bx2 + cx + d = y
(-1, 15) ---> a(-1)3 + b(-1)2 + c(-1) + d = 15
(0, 0) ---> a(0)3 + b(0)2 + c(0) + d = 0
(1, -5) ---> a(1)3 + b(1)2 + c(1) + d = -5
(2, 12) ---> a(2)3 + b(2)2 + c(2) + d = 12
Simplifying: -1a + 1b - 1c + 1d = 15
0a + 0b + 0c + 1d = 0
1a + 1b + 1c + 1d = -5
8a + 4b + 2c + 1d = 12
Solving these: a = 2 b = 5 c = -12 d = 0 [I used matrices.]
-----> f(x) = 2x3 + 5x2 - 12x
= x(2x2 + 5x - 12)
= x(2x - 3)(x + 4)
Solutions: x = 0 x = 3/2 x = -4
