dy/dx= (2x-2xy)/(x^2+2y)
find all points on the curve where x=2. show there is a horizontal tangent to the curve at one of those points.
To find all the points on the curve where x = 2, we need to substitute x = 2 into the given equation and solve for y.
Start with the given equation:
(dy/dx) = (2x - 2xy) / (x^2 + 2y)
Substitute x = 2:
(dy/dx) = (2(2) - 2(2)y) / (2^2 + 2y)
Simplify:
(dy/dx) = (4 - 4y) / (4 + 2y)
Now, we can find the points where x = 2 by equating dy/dx to zero and solving for y.
Setting (dy/dx) = 0, we have:
0 = (4 - 4y) / (4 + 2y)
Next, cross-multiply and solve for y:
0 = (4 - 4y)(4 + 2y)
0 = 16 - 16y + 8y - 8y^2
Rearrange to get a quadratic equation:
8y^2 - 8y + 16 = 0
Divide through by 8:
y^2 - y + 2 = 0
This quadratic equation does not factor, so we can apply the quadratic formula:
y = [ -(-1) ± √((-1)^2 - 4(1)(2))] / (2(1))
Simplifying further:
y = (1 ± √(-7)) / 2
Since the quadratic formula involves taking the square root of a negative number, we can conclude that there are no real solutions for y. Therefore, there are no points on the curve where x = 2.
Since there are no points where x = 2, we cannot find a point with a horizontal tangent on the curve when x = 2.