Question

I have zero idea how to do this http://imgur.com/2lsFiAN

Answer 1

lets write f(x) in terms of its roots

Answer 2

x = -4, -2, 2, 4
are the roots?

Answer 3

Any polynomial can be written in the form
f(x) = a * (x - r1)(x-r2)(x-r2)... (x-rn)

Answer 4

oh (I didn't know that)
ok
so the numbers I just said can be plugged in for the r1 r2 stuff?

Answer 5

right :)

Answer 6

also you might want to plug in the point (0,-3)

Answer 7

but first plug in the roots you know

Answer 8

oh i should mention, there is going to be multiplicity

Answer 9

f(x) = a * (x + 4) * ( x - 4) * (x - 2) * (x +2)

what about the x = 4 that does not pass through the x axis?
and where would I plug in (0, -3)

Answer 10

your root 4 , do you see how it doubles back, it is going to have multiplicity 2 or some even power

Answer 11

it does not cross the x axis

Answer 12

f(x) = a * (x + 4) * ( x - 4)^2 * (x - 2) * (x +2) - 3 ?

I suppose I can't just stick the -3 on the end like that ;-/

Answer 13

Let x = 0, f(x)=y=-3

Answer 14

$$\Large {
f(x) = a (x+4) (x+2)(x-2)( x - 4)^\color{red}{2}

}
$$

Answer 15

notice how the degree of this is now power 5, which makes sense, your graph has an odd degree poly.

Answer 16

then do I have to make f(0) = -3 true?

Answer 17

plug in zero for x, -3 for f(x) , and solve for `a`

Answer 18

-3 = a(4)(2)(-2)(16) ?

Answer 19

a = 3/256 ???

Answer 20

Yep.

Answer 21

so y(x) = (3/256)*(x+4)(x+2)(x-2)(x-4)^2 would be my answer?

Answer 22

Yep.

Answer 23

thanks

Answer 24

Np