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

To find the equation of the tangent line at the point where t=-1, we use the point-slope form of a line equation: y - y1 = m(x - x1), where (x1, y1) is the point on the curve and m is the slope of the tangent line.

From the given information, we know that the point on the curve when t=-1 is (1, 7), and the slope of the curve at that point is -1/2.

Substituting these values into the point-slope equation gives us:
y - 7 = (-1/2)(x - 1)

Simplifying, we can rewrite the equation in slope-intercept form (y = mx + b), as follows:

y - 7 = (-1/2)x + 1/2
y = (-1/2)x + 15/2

Therefore, the equation of the tangent line at the point where t=-1 is y = (-1/2)x + 15/2.