Determoine the cubic function with zeros -2, 3, and 4 and F(5) = 28
Since the zeros are listed... start with them : )
y= a(x+2)(x-3)(x-4)
now you have the F(5) = 28 so set each of your x's to 5 and your y to 28 and solve for "a" : )
So clearly f(x) = a(x+2)(x-3)(x-4)
also f(5) = 28, so
28 = a(7)(2)(1)
a = 2
f(x) = 2(x+2)(x-3)(x-4) , expand it if you need to, I like it better this way
@Reiny where'd you get the x-4?
To determine the cubic function with zeros at -2, 3, and 4, we can use the zero product property. The zero product property states that if a product of factors equals zero, then at least one of the factors must be zero.
So, if the zeros of the cubic function are -2, 3, and 4, we can write the factors of the function as:
(x - (-2))(x - 3)(x - 4) = 0
(x + 2)(x - 3)(x - 4) = 0
Now, let's multiply the factors together to get the cubic function in factored form:
(x + 2)(x - 3)(x - 4) = 0
Expanding this expression, we get:
(x^2 - x - 6)(x - 4) = 0
Now, let's distribute the (x - 4) term:
(x^2 - x - 6)(x) - (x^2 - x - 6)(4) = 0
x^3 - x^2 - 6x - 4x^2 + 4x + 24 = 0
x^3 - 5x^2 - 2x + 24 = 0
Therefore, the cubic function with zeros -2, 3, and 4 is:
f(x) = x^3 - 5x^2 - 2x + 24.
To verify if F(5) = 28, we substitute x = 5 into the cubic function:
f(5) = (5)^3 - 5(5)^2 - 2(5) + 24
f(5) = 125 - 125 - 10 + 24
f(5) = 14.
Thus, the cubic function f(x) = x^3 - 5x^2 - 2x + 24 satisfies the condition F(5) = 28.