Find the point(s) on the curve where the tangent is horizontal. Select the correct answer.
$$x = t ^ { 3 } - 3 t + 2 , y = t ^ { 3 } - 3 t ^ { 2 } + 2$$
A) $( 0 , - 0 )$
B) $( 0,0 ) , ( 2 , - 2 )$
C) $( - 1,1 ) , ( - 2 , - 2 )$
D) $( 0,0 ) , ( 4 , - 2 )$
E) None of these

Question

Quizplus · Accepted Answer

To find where the tangent is horizontal, we need to find where the derivative of y with respect to x is equal to zero (since the slope of the tangent is the derivative).

$$\frac { dy } { dx } = \frac { \frac { d y } { d t } } { \frac { d x } { d t } } = \frac { 3 t ^ { 2 } - 6 t } { 3 t ^ { 2 } - 3 } = 0$$

Solving for t, we get t=0 and t=2.

Plugging these values of t back into the equation of the curve, we get the points where the tangent is horizontal:

$$t=0\implies (2,2)$$ 
$$t=2\implies (4,-2)$$

Therefore, the answer is D, which lists both of these points.

Find the Point(s) on the Curve Where the Tangent Is x=t3−3t+2,y=t3−3t2+2x = t ^ { 3 } - 3 t + 2 , y = t ^ { 3 } - 3 t ^ { 2 } + 2x=t3−3t+2,y=t3−3t2+2

Find the Point(s) on the Curve Where the Tangent Is $x = t ^ { 3 } - 3 t + 2 , y = t ^ { 3 } - 3 t ^ { 2 } + 2$