+4609 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 \(f\)?
+128406 f(-1) = 15
f(0) = 0
f(1) = -5
f(2) = 12
A cubic has the form ax^3 + bx^2 + cx + d
Since (0,0) is on the graph, then d must equal 0
So.....we have this system
-a + b - c = 15
a + b + c = -5
8a + 4b + 2c = 12
-a + b - c = 15
a + b + c = -5 add these and we get that 2b = 10 ⇒ b = 5
And
8a + 5(4) + 2c = 12
8a + 2c = -8 using this and -a + b - c = 15 we have that
8a + 2c = -8
-a + 5 - c = 15
8a + 2c = -8 (1)
-a - c = 10 ⇒ -2a - 2c = 20 (2) add (1) and (2)
6a = 12 ⇒ a = 2
And using - a + b - c = 15
2 + 5 + c = -5 ⇒ c = - 12
So.... the function is
2x^3 + 5x^2 - 12x .......to find the x intercepts.....set this to 0
2x^3 + 5x^2 - 12x = 0 factor
x (2x^2 + 5x - 12) = 0
x (2x - 3) (x + 4) = 0
Setting the factors to 0 and solving for x we get that the x intercepts are
x = -4 , x = 0 and x = 3/2 = 1.5
