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?

Guest Aug 26, 2019

#1**+2 **

Since f(0) = 0,,,,then x = 0 is one root

So.....we have the form ax^3 + bx^2 + cx + d where d = 0

So we know that

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)

Add (1) and (2) and we have that 2b = 10 ⇒ b = 5

Multiply (1) by 8 and add to (3) and we have that

12b - 6c = 132 sub 5 in for b and we have

12(5) - 6c = 132

60 - 6c = 132 subtract 60 from both sides

-6c = 72 divider both sides by -6

c = -12

And using (2) we can find a as

a + 5 - 12 = -5

a - 7 = -5 add 7 to both sides

a = 2

So.....the polynomial is 2x^3 + 5x^2 - 12x

Set to 0 and factor

x (2x^2 + 5x - 12) = 0

x (2x - 3) ( x + 4) = 0 set each factor to 0 and solve for x

x = 0 2x - 3 = 0 x + 4 = 0

x = 0 x = 3/2 x = -4 these are the x intercepts

Here is a graph : https://www.desmos.com/calculator/o040mctwqm

CPhill Aug 26, 2019