Question

find the points on the curve where the tangent is horizontal x= t^3 -3t y=t^3 - 3t^2

Answer 1

i already got the dy/dx which is 3t^2 -6t / 3t^2 - 3

Answer 2

but i have no clue what to do next

Answer 3

My answer was very verbose so I decided to concise it.
Ok so we have:
$$
\begin{align*}
x &= t^3 - 3t& y &= t^3 - 3t^2\\
\frac{dx}{dt} &= 3t^2 - 3 &\frac{dy}{dt} &= 3t^2 - 6t
\end{align*}
$$So, the slope of our tangent is:
$$
\frac{dy}{dx} = \frac{\;\; \frac{dy}{dt}\;\; }{ \frac{dx}{dt} } = \frac{3t^2 - 6t}{3t^2 - 3} \\
$$Since division by 0 is not defined, we have to check for what \(t\) values our slope is defined:
$$
3t^2 - 3 \neq 0 \implies 3t^2 \neq 3 \implies t^2 \neq 1\\
t \neq \pm \sqrt{1} \implies t \neq \pm 1
$$Now we want to check when our tangent is horizontal, which means it has slope of 0:
$$
\frac{dy}{dx} = 0 \implies \frac{3t^2 - 6t}{3t^2 - 3} = 0 \\
3t^2 - 6t = 0 \\
t_{1,2} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot 0}}{2 \cdot 3} = \frac{6 \pm \sqrt{36}}{6} = \frac{6\pm6}{6} \\
t_1 = \frac{6 - 6}{6} = \frac{0}{6} = 0 \qquad t_2 = \frac{6 + 6}{6} = \frac{12}{6} = 2
$$Both the values are possible for \(t\) as they don't cause division by 0.
Now we have to find the \(x,y\) values for these \(t\) values in order to get the points on the curve
$$
\begin{array}{llll}
t = 0 \implies x = 0^3 - 3\cdot 0 = 0& &y=0^3 - 3\cdot 0^2 = 0 \\
P_1 = (0,0) \\
t = 2 \implies x = 2^3 - 3\cdot 2 = 8-6 = 2 & &y=2^3 - 3\cdot 2^2 = 8 - 12 = -4 \\
P_2 = (2, -4)
\end{array}
$$
I also made an animated graph here:
https://www.desmos.com/calculator/utagyqgcm5
You can see that the point \((x,y)\) changes its position as \(t\) changes.
It is analogous to a position of a pen drawing the curve over time, I think it's nice =)
The orange points are \(P_1\) and \(P_2\). Also notice that \(P_2\) is where the graph intersects with itself, because you get the same \(x,y\) coordinates for t=-1 as well.

Hope it helps \( \Large ☺ \)