Answer:
The parabola has a horizontal tangent line at the point (2,4)
The parabola has a vertical tangent line at the point (1,5)
Step-by-step explanation:
Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.
-To obtain horizontal tangent lines, the condition is:
(The slope is zero)
--To obtain vertical tangent lines, the condition is:
(The slope is undefined, therefore the denominator is set to zero)
Derivating respect to x:
=
Solving for dy/dx:

Applying the first conditon (slope is zero)

Solving for y (Adding 2x+4, dividing by 2)
y=x+2 (I)
Replacing (I) in the given equation:

Replacing it in (I)
y=(2)+2
y=4
Therefore, the parabola has a horizontal tangent line at the point (2,4)
Applying the second condition (slope is undefined where denominator is zero)
2y-2x-8=0
Adding 2x+8 both sides and dividing by 2:
y=x+4(II)
Replacing (II) in the given equation:

Replacing it in (II)
y=1+4
y=5
The parabola has vertical tangent lines at the point (1,5)