Question

for the parametric curve defined by x=3-2t^2 and y=5-2t ...sketch the curve using the parametric equation to plot of the point. use an arrow to indicate the direction of the curve for o<t<1. Find an equation for the line tangent to the curve at the point where t=-1.

x=3 - 2t^2 , so dx/dt = -4t
y= 5 - 2t , so dy/dt = -2

it follows that (dy/dt)÷(dx/dt)
=dy/dt
=-2/-4t
= -1/2 when t=-1

by assigning different values to t, you can generate x and y values, thus a few points.
It should be obvious quickly that your would get a parabola with vertex at (3,5), axis of symmetry y = 5, opening to the left

when t=-1, the point is (1,7) and slope =-1/2

use that information and y = mx + b to find the tangent equation