Answer
Points of horizontal tangency: $(2,-2)$ and $(4,2)$.
Point of vertical tangency: $(\frac{7}{4},\frac{-11}{8})$
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:
$x=t^2-t+2=(1)^2-1+2 =2$
$y=t^3−3t=1^3−3∗1=1−3=−2$
Therefore, one point of horizontal tangency is (2,−2)
Substitute the values at t = -1:
$x=t^2-t+2=(-1)^2-(-1)+2 = 4$
$y=t^3−3t=(−1)^3−3(−1)=−1+3=2$
Therefore, the other point of horizontal tangency is (4,2)
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The point of vertical tangency will occur when dx is equal to 0:
$x = t^2 - t + 2$
$dx = (2t - 1)dt$
dt can't be equal to 0, so: $2t - 1 = 0$
$2t = 1$
$t = 1/2$
Solve for "x" and "y" when $t= \frac{1}{2}$
$x = t^2 -t + 2 = (1/2)^2 - (1/2) + 2$
$x = (1/4) - (1/2) + 2 = \frac{7}{4}$
$y = t^3 - 3t$
$y = (1/2)^3 - 3(1/2)$
$y = (1/8) - (3/2) = \frac{-11}{8}$
Therefore, the point of vertical tangency is $(\frac{7}{4},\frac{-11}{8})$
