Find the equation of the Tangent Line to the given set of Parametric Equations at given point.

Question

I'm looking for validation for my answer to this question.

Parametric Equations: $x = t^2 + 2t + 1 , y = t^3 + 7t^2 + 8t, t = -1$

For this problem I used the Point-Slope-Form formula.

myAnswer:$ y = 19/2X + 17$

Answer 1

HINT

at $ t=-1 , (x_1,y_1)= (0,-2)$

$$ m=\dfrac{dy}{dx}= \dfrac{dy/dt}{dx/dt} $$

what is Point-Slope formula?

EDIT1:

slope calculation ( error corrected)

$$ \dfrac{y+2}{x}=\dfrac{3 t^2+14 t +8}{2t+2}=\dfrac{3}{0}$$

Answer 2

The equation of the tangent at $t$ is

$$\frac{x-f(t)}{f'(t)}=\frac{y-g(t)}{g'(t)}.$$

With the given,

$$\frac{x-0}{0}=\frac{y+2}{3}.$$

As the left denominator is zero, the tangent must be vertical, and its equation reduces to

$$x=0.$$