Answer
Point of horizontal tangency: $(\frac{-1}{2},\frac{-9}{4})$
Work Step by Step
The point of horizontal tangency will occur when $dy/dx = 0$, therefore when $dy$ is zero, in this case:
$y = t^2 + 3t$
$dy = (2t + 3)dt$
dt can't be 0, so: $2t + 3 = 0$: $t = \frac{-3}{2}$
Substitute the values at t = $\frac{-3}{2}$:
$x = t + 1 = \frac{-3}{2} + 1 = \frac{-1}{2}$
$y = t^2 + 3t= (\frac{-3}{2})^2 +3\frac{-3}{2}= \frac{9}{4} - \frac{9}{2} = \frac{9-18}{4} = \frac{-9}{4}$
Therefore, the point of horizontal tangency is $(\frac{-1}{2},\frac{-9}{4})$
The point of vertical tangency will occur when $dx$ is equal to 0:
$x = t+1$
$dx = 1dt$
1dt can't be equal to 0, so there is no vertical tangency point.
