Need help with a Polynomial Function Graph given the points

Nov 2019
6
0
California
I am really struggling with polynomial function graphs and was assigned the points and I am lost as too how to reverse engineer it to get the function. Image below are the points and graph shape.

Screen Shot 2019-11-17 at 7.09.28 PM.png
 

Debsta

MHF Helper
Oct 2009
1,313
600
Brisbane
Are you given the coordinates of the local maximum turning point?

Also, am I correct in assuming that the points (-4,0) and (2,0) are minimum turning points?
 
Jun 2013
1,112
590
Lebanon
in other words,

(-4,0) , (2,0), and (0,16)

are three points on the graph of what looks like a fourth degree polynomial

is that it?
 
Nov 2019
6
0
California
The function like f(x) = anX^n + a n-1 x^n-1 + ... + a1x + a0. Essentially the function that will replicate it on something like desmos.com
 
Nov 2019
6
0
California
Are you given the coordinates of the local maximum turning point?

Also, am I correct in assuming that the points (-4,0) and (2,0) are minimum turning points?
correct. Those were the only points given.
 

Debsta

MHF Helper
Oct 2009
1,313
600
Brisbane
There are a couple of ways of doing this.

The easiest way:

Has 3 turning points, so possibly a fourth degree polynomial.

Because of the nature of the 2 (and only 2) roots (ie occurring at TPs sitting on the x-axis) then can you say what the factorised form of the function will look like?
 
Nov 2019
6
0
California
There are a couple of ways of doing this.

The easiest way:

Has 3 turning points, so possibly a fourth degree polynomial.

Because of the nature of the 2 (and only 2) roots (ie occurring at TPs sitting on the x-axis) then can you say what the factorised form of the function will look like?
I came up with -2x^{2}-4x+16 which when put into desmos is nowhere near correct because the line doesn't go up after the two roots.
 

Debsta

MHF Helper
Oct 2009
1,313
600
Brisbane
No that's a parabola (ie a second degree polynomial).

Do you know that if the graph has a TP sitting on the x-axis at (2, 0) then one of the factors is (x-2) raised to an even power?
Same for the point (-4,0)
 
Nov 2019
6
0
California
No that's a parabola (ie a second degree polynomial).

Do you know that if the graph has a TP sitting on the x-axis at (2, 0) then one of the factors is (x-2) raised to an even power?
Same for the point (-4,0)
I think I am on the right path. Right now I got (x)^{4}\(x-2)^{2}(x+4)+16. Still not correct but closer
 

Debsta

MHF Helper
Oct 2009
1,313
600
Brisbane
Easier to work with it in factorised form.
Because it is a fourth degree polynomial with factors of (x-2) raised to an even power and (x+4) raised to an even power then
\(\displaystyle f(x) = a(x-2)^2(x+4)^2\) to give a fourth degree polynomial (when expanded out).

Your job now is to find a such that the graph goes through (0, 16).