The polynomial has degree 3. If f(-1) = 32, f(0) = 0, f(1) = -24, ,and f(2) = -28, then what are the x-intercepts of the graph of f?
+128407 We have the form P(x) = ax^3 + bx^2 + cx^2 + d
Since f(0) = 0 .....then d = 0
So we actually have this system of equations
a(-1)^3 + b(-1)^2 + c(-1) = 32
a(1)^3 + b(1)^2 + c(1) = -24
a(2)^3 + b(2)^2 + c(2) = -28 simplify these
-a + b - c = 32 (1)
a + b + c = -24 (2)
8a + 4b + 2c = -28 (3)
Add (1) and (2)
2b = 8
b = 4
Sub this into (2) and (3)
a + 4 + c =-24 → a + c = -28 (4)
8a + 4(4) + 2c = -28 → 8a + 2c = -44→ 4a + c = -22 → -4a - c = 22 (5)
Add (4) and (5)
-3a = -6
a = 2
Using (4)
2 + c = -28
c = -30
Our polynomial is
P(x) = 2x^3 + 4x^2 - 30x
Setting this to 0 and dividing through by 2, we have that
x^3 + 2x^2 -15x = 0 factor
x (x^2 + 2x - 15) = 0
x ( x -3) (x + 5) = 0
Setting each factor to 0 and solving for x, the intercepts are
x = 0 x = 3 and x = -5
See the graph here : https://www.desmos.com/calculator/zrdafdxpea
