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?
if someone could help me out that'dbe great, thanks!
+118608 Let me ask YOU some questions. Just to help you learn.
One of them can be seen immediately. What one is that?
How many (at most) can there be?
+128473 A 3rd degree polynomial has the form ax^3 + bx^2 + cx + d
Since f(0) = 0 then x = 0 is one of the y intercepts and d = 0
So....we can solve for a, b and c as follows :
a(-1)^3 + b(-1)^2 + c(-1) = 15 ⇒ -a + b - c = 15 (1)
a(1)^3 + b(1)^2 + c(1)= -5 ⇒ a + b + c = -5 (2)
a(2)^3 + b(2)^2 + c(2) = 12 ⇒ 8a + 4b + 2c = 12 (3)
Adding (1) and (2) we have that 2b = 10 ⇒ b = 5
Sub this into each equation for b and we have that
-a +5 - c = 15 ⇒ -a - c = 10 (4)
a +5 + c = -5 ⇒ a + c = -10 (5)
8a + 4(5) + 2c = 12 ⇒ 8a + 2c = -8 (6)
Multiply (5) by -2 and add it to (6) and we get that 6a =12 ⇒ a = 2
Using (5) to find c we have that 2 + c = -10 ⇒ c =-12
So....the polynomial is f(x) = 2x^3 + 5x^2 - 12x set to 0 and factor and we get that
(2x - 3) (x + 4 ) = 0
Setting each factor to 0 and solving for x and we get the other two x intercepts of x = 3/2 and x = -4
See the graph here : https://www.desmos.com/calculator/ef2xuaesrm
