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.

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?

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?

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)

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)

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).