Question

what is the equation of a polynomial function that has zeros at -4,-2, and 1. and that passes through the point (-3,12)

Answer 1

\[x^3+5 x^2+2 x-8\]

Answer 2

well, if we use whats given and construct one with what we should know; it will tend to just coaless into the proper form

Answer 3

it has 3 zeros; meaning

0*0*0 = 0
(x+4)(x+2)(x-1)=0

now to antifoil lol

(x+4)
(x+2)
-----
x^2 +4x
+2x +8
------------
x^2 +6x +8
x-1
------------
x^3 +6x^2 +8x
-x^2 -6x -8
----------------
x^2 +5x^2 +2x -8

x^2 +5x^2 +2x -8; is the base function, which we need to modify to hit the given point, so lets start by seeing how far off we are

Answer 4

when (x=-3,y=12)

x^3 +5x^2 +2x -8 = y

(-3)^3 +5(-3)^2 +2(-3) -8 = 12

-27 +45 -6 -8 = 12

-27 +45 -14 = 12

-41 +45 = 12

4 = 12 ; the only modification we can make is a scaled modification, in other words multiply or divde

4s = 12 when s = 3

y = 3(x^3 +5x^2 +2x -8)

y = 3x^3 +15x^2 +6x -24

Answer 5

Or in other words DHASHNI was wrong.

Answer 6

Dhashni provide the base model :)

Answer 7

we had to add the customized headers and pleather front bench seating

Answer 8

thanks man, helped alot

Answer 9

A slightly more compact way of dealing with 'passes through (-3,12) is as follows.

The polynominal must have the form for some number A

p(x) = A(x+4)(x+2)(x-1), as -4, -2 and 1 are roots.

So if (-3,12) lies on the graph, p(-3) = 12

I.e., A.1.-1.-4 = 12, hence 4A = 12, A = 3

So the polynominal is p(x) = 3(x+4)(x+1)(x-1)

Answer 10

ANTIFOIL!

Answer 11

.... whenever i do math my knuckles bleed :)