Answer
There are 2 horizontal tangency points: $(5,-2)$ and $(3,2)$
Work Step by Step
The point of horizontal tangency will occur when $dy/dx = 0$, therefore when $dy$ is zero, in this case:
$y = t^3 - 3t$
$dy = (3t^2 - 3)dt$
dt can't be equal to 0, so: $3t^2 - 3 = 0$: $3t^2 = 3$ : $t^2 = 1$ : $t = ±1$
Substitute the values at $t = 1$:
$x = t + 4 = 1 + 4 = 5$
$y = t^3 - 3t = 1^3 - 3*1 = 1 -3 = -2$
Therefore, one point of horizontal tangency is $(5,-2)$
Substitute the values at t = -1:
$x = t + 4 = -1 + 4 = 3$
$y = t^3 - 3t = (-1)^3 -3(-1) = -1 +3 = 2$
Therefore, the other point of horizontal tangency is $(3,2)$
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The point of vertical tangency will occur when $dx$ is equal to 0:
$x = t+4$
$dx = 1dt$
1dt can't be equal to 0, so there is no vertical tangency point.
