Find an equation for the line tangent to the parametric curve at the given value of t

$x=2t-1,y=3t+5,t=-1$

The parametric curve is$x=2t-1,y=3t+5,t=-1$

Consider the parametric curves $x=2t-1,y=3t+5$at $t=-1$.

The objective is to find the equation of a tangent line for the given parametric equations.

The formula to find the tangent line equation is$y-y_1=m\left(x-x_1\right)$.

First find the slope of the parametric curves by finding the derivative of the parametric curves.

For that we use the formula $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

Now take the parametric equation $x=2t-1$.

Differentiate the curve with respect to t.

Then,

$$\begin{gathered} \frac{dx}{dt}=\frac{d(2t-1)}{dt}\text{since}x=2t-1 \\ \frac{dx}{dt}=2·\frac{dt}{dt}-\frac{d1}{dt} \\ \frac{dx}{dt}=2·1-0\text{since}\frac{d1}{dt}=0 \\ \frac{dx}{dt}=2 \end{gathered}$$

Now take the parametric equation $y=3i+5$.

Differentiate the curve with respect to t.

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}(3t+5)\text{since}y=3t+5 \\ \frac{dy}{dt}=3\frac{d}{dt}t+\frac{d}{dt}5 \end{gathered}$$

The slope of the parametric equation is$\frac{dy}{dx}=m=\frac{3}{2}$.

The point $(x,y)$When $t=-1$is,

$$\begin{gathered} (x,y)=(2t-1,3t+5) \\ (x,y)=\left(2\right(-1)-1,3(-1)+5)[\text{since by substituting}t=-1] \\ (x,y)=(-3,2)\text{simplify} \end{gathered}$$

Now the slope point formula is $y-y_1=m\left(x-x_1\right)$.

The point is $(-3,2)$ and the slope is $m=\frac{3}{2}$.

$$\begin{gathered} y-2=\frac{3}{2}(x-(-3\left)\right)\left[\text{since}y-y_1=m\left(x-x_1\right)\right] \\ y-2=\frac{3}{2}(x+3) \end{gathered}$$

Therefore, the tangent line at $t=-1$ for the parametric equations is $y-2=\frac{3}{2}(x+3)$