Answer
Horizontal tangents at: $(0,0)$ and $(2,-4)$
Vertical tangents at $(-2,-2)$ and $(2,-4)$
Work Step by Step
1. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$:
$\frac{dy}{dt} = \frac{d(t^3-3t^2)}{dt} = 3t^2 - 6t$
$\frac{dx}{dt} = \frac{d(t^3-3t)}{dt} = 3t^2 - 3$
Therefore, $\frac{dy}{dx} = \frac{3t^2 - 6t}{3t^2-3} = \frac{3(t^2-2t)}{3(t^2-1)} = \frac{t^2-2t}{t^2-1}$
2. There will be a horizontal tangent when $t^2-2t = 0$ and $t^2-1 \neq 0$:
$t^2 - 2t = 0$
$t(t-2) = 0$
So:
$t = 0$
or
$t - 2 = 0 \longrightarrow t = 2$
- Check if $t^2-1 \neq 0$ for these values:
$(0)^2 - 1= -1$
$(2)^2-1 = 3$
- They are both valid.
3. There will be a vertical tangent when $t^2-1 = 0$ and $t^2-2t \neq 0$:
$t^2 - 1 = 0$
$t^2 = 1$
$t = 1$ or $t = -1$
- Check if $t^2-2t \neq 0$
$(1)^2-2(1) = 1 - 2 = -1$
$(-1)^2 - 2(-1) = 1 + 2 =3$
- They are both valid.
4. Find the points:
$t = 0$:
$x = (0)^3-3(0)= 0$
$y = (0)^3-3(0)^2= 0$
$t = 2$:
$x = (2)^3-3(2)= 2$
$y = (2)^3-3(2)^2= -4$
- Thus, there are horizontal tangents at $(0,0)$ and at $(2,-4)$.
$t = 1$:
$x = (1)^3-3(1)= -2$
$y = (1)^3-3(1)^2= -2$
$t = -1$:
$x = (-1)^3-3(-1)= 2$
$y = (-1)^3-3(-1)^2= -4$
- Thus, there are vertical tangents at $(-2,-2)$ and at $(2,-4)$.
