State the points on the curve where the tangent is horizontal AND vertical

Question

x = t^3 - 3t

y = t^2 -3

I tried the problem and this is what I got:

Horizontal: (0,-3)

Vertical: (2,-2) and (-2,-2)

Also I'll throw this one in since I have no one to check my work:

Find the equation of a tangent line to the curve at the point t=1

x = 1+4t-t^2

y = 2-t^3

For my equation I got:

y=-3x/2 + 7

Answer 1

Hint. One may recall that

$$ x'(t_0)\neq 0,\quad y'(t_0)= 0 $$

gives an horizontal tangent at the point $(x(t_0),y(t_0))$ and that

$$ y'(t_0)\neq 0,\quad x'(t_0)= 0 $$

gives an vertical tangent at the point $(x(t_0),y(t_0))$.

Here one gets $$ x'(t_0)=3t^2-3, \quad y'(t_0)=2t. $$

Can you take it from here?

Answer 2

$$\frac{dx}{dt}=3t^{2}-3=0=>t=^{+}_{-}1$$ for $t=1$ we have $$x=1^{3}-3\cdot1=-2$$ $$y=1^{2}-3=-2$$ this equal to point $$(-2,-2)$$ for $t=-1$ doing the same procedure, we have the point (2,-2). Now for the vertical: $$\frac{dy}{dt}=2t=0=>t=0$$ for t=0, we have: $$x=0^{3}-3\cdot0=0$$ and $$y=0^{2}-3=-3$$ wich is the point (0,-3). Now the second question we need the slope m of the curve first, for this we have the next $m=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$,so. $$\frac{dy}{dt}=-3t^{2}$$ for $t=1$ we have $$\frac{dy}{dt}=-3$$ now we calculate for $\frac{dx}{dt}$ $$\frac{dx}{dt}=4-2t$$ for t=1 we have $$\frac{dx}{dt}=4-2=2$$ then the slope m are $$m=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=-\frac{3}{2}$$, now calculate point $(x(1),y(1))$ $$x(1)=1+4-1=4=>y(1)=2-1=1 =>[x(1),y(1)]=[4,1]$$ we have the slope and the point, we can calculate now finally: $$y-1=-\frac{3}{2}(x-4)<=>y=\frac{3}{2}x+7$$ in conclution your answer are good.