#### Answer

$y=-3x+2$

#### Work Step by Step

The question asks us to find the equation of the tangent of the function
$f(x)=x -2x^2$ at the point $(1,-1)$
Given that we are looking for the equation of a tangent, the first reasonable step would be to find the gradient at the points given, which can be found using the derivative function.
Using the power rule:
$f'(x)=1-4x$
$f'(1)=-3$
Thus, the gradient of that function at the value x=1 is -3.
Using this information, we can find the intercept value of the tangent line.
The equation of the gradient is given by:
$\frac{y_{1}-y_{0}}{x_{1}-x_{0}}=m$
We can now use the points $(1,-1)$ and $(0,c)$ (where c is the y-intercept).
$\frac{-1-c}{1-0}=-3$
Solving for c gives us $c=2$
Thus, the equation of the tangent at $(1,-1)$ is $y=-3x+2$