Question

What is the slope of `f(t) = (t-2t^2,t)` at `t =-1`?

Answer 1

Slope at `t=-1` is `1/5`

Explanation:

The slope of a function `f(x)` is given by its derivative `f'(x)`.

In a parametric equation of type `f(t)=(x(t),y(t))`, `f'(t)=((dy)/(dt))/((dx)/(dt))`

As `(dy)/(dt)=1` and `(dx)/(dt)=1-4t`

`f'(t)=1/(1-4t)`

and slope at `t=-1` is `1/5` and this at `(-3,-1)`

graph{(x-y+2y^2)(5y+5-x-3)=0 [-5.813, 4.187, -2.68, 2.32]}