First look at the graph — looks super nonlinear and pretty exponential
Use the point (0,4).
We can eliminate answers (2) and (3), as if we substitute h=0,  the output would be 6/5 and 4.2 as opposed to the 4 for the y value we are looking for.
Now we can use the point (1,8) to see if y = 8 when x = 1
For answer (4), if we substitute in x = 1, we get 2/3(1)^3 – 1^2 + 3(1) + 4 = 2/3 – 1 + 3 + 4≠ 8
Try answer (1): using order of operations: 4(2)^1 = 4(2) = 8
4(2)^2 = 16 and 4(2)^3 = 32.  That’s it –2 points
You can also use Y= and check the table for a match: