Question

Find a third-degree polynomial function such that f(0) = -24 and whose zeros are -1, 2, and -3. Using complete sentences, explain how you found it.

Answer 1

alright for this question zeroes are basically the x-intercept so just think of it this way you are given a quaratic equation say ax^2 +bx +c =f(x) to find the zeroes for this function we will let f(x) =0 and then try to factor that equation down to something like (x-c)(x+d) or something like that and our x intercepts are x=c and x=-d

so in this case we are already give x= -1 x=2 x= -3 and now we are just working back wards ok so our factored form is (x+1)(x-2)(x+3)=f(x) and now we just expand the left hand side to (x^2 -x -2)(x+3) = f(x) and that is now f(x) =x^3 +3x^2 -x^2 +3x -2x -6 which is if we add like terms

f(x) = x^2 +x^2 +x-6 :) and but now we need to satisfy when x=0 y must =-24

Answer 2

f(x) = x^3 + x^2 +x -6 i forgot to write 3

Answer 3

f(x) = (x+1)(x-2)(x+3)

From hamza_b23's explanation;
We know that this function has its y-intercept at (0, -6)

To get (0, -24) as the equation's y-intercept
you simply stretch to function, so you change the slope, -24/-6 = 4
The slope should be 4 if you want to satisfy when x=0, y=-24
So the final equation would be

\[\large f(x) = 4(x+1)(x-2)(x+3)\]

Answer 4

Then you have to expand it out,
\[\large\ f(x) = 4(x+1)(x-2)(x+3)\]
\[\large\ f(x) = 4(x^2-x-2)(x+3)\]
\[\large\ f(x) = 4(x^3 -x^2-2x+3x^2-3x-6)\]
\[\large\ f(x) = 4(x^3+2x^2-5x-6)\]
\[\large\ f(x) = 4x^3+8x^2-20x-24\]

Answer 5

you don't necessarily need to expand it.

Answer 6

Oh, but the general form is much cooler :D

Answer 7

in general, if you have an n'th degree function with roots (zeros) of \(x=r_1, x=r_2, ..., x=r_n\), then you can write it as:\[f(x)=k(x-r_1)(x-r_2)...(x-r_n)\]

Answer 8

where k is some constant.

Answer 9

zepp: depends how you measure "coolness" :)

Answer 10

Oh, constant.. I wrote slope :s

bah, I just feel like expanding it, that's all :P

Answer 11

each to their own.... :D