Answer
$$\frac{(x-4)^2}{16}+\frac{(y-2)^2}{4}=1$$
Work Step by Step
If this graph was centered at the origin, we would see that it would have y-intercepts at $\pm 2$ and x intercepts at $\pm 4$
Thus the equation would be,
$$\frac{x^2}{16}+\frac{y^2}{4}=1$$
We then shift it right 4 and up 2 to get...
$$\frac{(x-4)^2}{16}+\frac{(y-2)^2}{4}=1$$
