How do you find the critical points for #(x^2)/36 + (y^2)/16 =1#?
1 Answer
Dec 13, 2015
Critical points:
Explanation:
The general form for an ellipse is
with
Converting the given equation to this form, we have
Giving us critical points on the Y-axis of
and critical points on the X-axis of
graph{x^2/36+y^2/16=1 [-8.89, 8.89, -4.444, 4.44]}