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

Nov 21, 2017

Slope at $t = - 1$ is $\frac{1}{5}$

#### Explanation:

The slope of a function $f \left(x\right)$ is given by its derivative $f ' \left(x\right)$.

In a parametric equation of type $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$, $f ' \left(t\right) = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

As $\frac{\mathrm{dy}}{\mathrm{dt}} = 1$ and $\frac{\mathrm{dx}}{\mathrm{dt}} = 1 - 4 t$

$f ' \left(t\right) = \frac{1}{1 - 4 t}$

and slope at $t = - 1$ is $\frac{1}{5}$ and this at $\left(- 3 , - 1\right)$

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