Question

Help please
Find a formula for the polynomial of least degree that is graphed below. The graph has x-intercepts −4,0,4 and goes through the points (−3,9) and (3,9).
Given graph:
http://img819.imageshack.us/img819/6255/nkizi2138set03a2polynom.png

Answer 1

The link is not clickable for some reason, so you will haev to copy paste it into the browser

Answer 2

http://img819.imageshack.us/img819/6255/nkizi2138set03a2polynom.png

Answer 3

zeros of the function are -4, 0, 4 so you will have (x+4)(x)(x-4)

Answer 4

Hmm what Marina said is wrong. I tried entering that, and i tried foiling them which comes out to be x^(3)-16x. Both come out to be wrong

Answer 5

It is not wrong. It is not done yet. I want you to understand th eprocess.

Answer 6

On the graph it bounce at x=0, so it will be x^2

Answer 7

why would it bounce? Doesnt it have to have an even power to bounce? x^(3)-16x, would not the 0 pass through?

Answer 8

Lets use form f(x)= a(x+4)(x)^2(x-4) and point (-3,9)
x=-3 f(-3)=a(-3+4)(-3)^2(-3-4)=9
Find out value of a,

Answer 9

out it into equation and try again.

Answer 10

wait so plug in the 9 into X? or A

Answer 11

y=ax^2(x+4)(x-4) Plugging in x=3 and y=9 you get 9=a(9)(7)(-1) so a=-1/7 and plugging in x=-3 and y=9 you get a=-1/7. So the final equation is -1/7(x^2)(x+4)(x-4)

Answer 12

-1/7(x^(2))(x+4)(x-4) worked ^^

Answer 13

thank you :D

Answer 14

Good do you get the concept of using a to help write the equation, and the jumps are x^2.

Answer 15

honestly, this is in the chapter, but the math tests we take dont have these questions..however the homework does. So I don't know them very well.
As for the jumps, if we had a (x-4)(x)(x+4), would not x end up being x^(3) which means it would not bounce?

Answer 16

You are right to think this, because it is just a cubic because it has 3 x's but to graph it more easily just leave it (x-4)(x)(x+4). It doesn't jump because there are no two same x-intercepts but in the equation that you gave in the graph, 0 is a repeated factor because of the jump at (0,0)

Answer 17

Ah I see. You would have to look at the graph.

Answer 18

yup.

Answer 19

you are welcome. It is so rewarding to help out friendly and patient person.