Find all points (if any) of horizontal and vertical tangency to the curve $x = t + 6 , y = t ^ { 3 } - 27 t$
A) horizontal tangents: $( 9 , - 54 ) , ( 3,54 )$, vertical tangent: none
B) horizontal tangent: none, vertical tangents: $( 9 , - 54 ) , ( 3,54 )$
C) horizontal tangents: $( 9,54 ) , ( 3,54 )$, vertical tangent: none
D) horizontal tangent: none, vertical tangents: $( 9,54 ) , ( 3,54 )$
E) horizontal tangent: none, vertical tangent: none

Question

Quizplus · Accepted Answer

To find points of horizontal and vertical tangency, we first differentiate $x$ and $y$ with respect to $t$. We get $dx/dt = 1$ and $dy/dt = 3t^2 - 27$. For horizontal tangents, we set $dy/dt = 0$ to find $t$ values, which gives $3t^2 - 27 = 0$, or $t^2 = 9$, so $t = \pm 3$. Substituting $t = 3$ and $t = -3$ into $x = t + 6$ and $y = t^3 - 27t$, we get the points $(9, -54)$ and $(3, 54)$ respectively. For vertical tangents, we would need $dx/dt = 0$, but since $dx/dt = 1$ for all $t$, there are no vertical tangents.

Find All Points (If Any) of Horizontal and Vertical Tangency x=t+6,y=t3−27tx = t + 6 , y = t ^ { 3 } - 27 tx=t+6,y=t3−27t

Find All Points (If Any) of Horizontal and Vertical Tangency $x = t + 6 , y = t ^ { 3 } - 27 t$