Question

use differentation 4x^2+y^2-8x+4y+4=0 vertical and horizontal tangent

Answer 1

Okay.
To find the vertical tangent, treat y as a constant and take the derivative of the equation with respect to x.
You will get:
8x -8 = 0
x = 1.
This is saying that you have a a vertical tangent (which simply means a line parallel to the y-axis ) at x=1.
Similarly,
To find the horizontal tangent, treat x as a constant and take the derivative of the equation with respect to y:
2y+4=0
y = -2

Answer 2

thank you