The calculus way to do this is to differentiate 4x^2−24x+4y^2−32y=−84 implicitly with respect to x, then solve for y' = dy/dx:
8x -24 +8yy' -32y' = 0
i.e. y' = (3-x)/(y-4).
Setting y' to 0 gives x = 3, which is the value of x where the slope of the tangent line to the curve is zero. Let's find the values of y corresponding to x = 3. Set x = 3 in the original equation then solve for y (you will use the quadratic equation to solve for y here):
4(3)^2−24(3)+4y^2−32y=−84
y = 2 or 6
Therefore, the curve has horizontal tangents at the given point (3,6), and the second point (3,2).
But the geometric way to solve this problem is much quicker! In the equation 4x^2−24x+4y^2−32y=−84, simply group x and y terms on the left-hand side, then complete the square in x and y, to get:
(x - 3)^2 + (y-4)^2 = 2^2
Now, recognize that this is just the circle with center (3,4) with radius = 2. The horizontal tangents must occur at the topmost and bottommost points of the circle, that is, the points (3,6) and (3, 2). Short and sweet!
