What is the equation of the tangent line of f(x)=6x−x2 at x=-1?

Explanation:
We are given
${\displaystyle f\left(x\right)=6x-x^2}$
To find the equation of the tangent line, we need to: find the slope of the tangent line, obtain a point on the line, and write the tangent line equation.
To find the slope of the tangent line, we take the derivative of our function.
f'(x)=6-2x
Substituting our point x=-1
f'(-1)=6-2(-1)=6+2=8
Now that we have our slope, we need to find a point on the line. We have an x-coordinate, but we need a f(x) too.
${\displaystyle f(-1)=6(-1)-{(-1)}^2=-6-1=-7}$
So the point on the line is (-1, -7).
With a slope and a point on the line, we can solve for the equation of the line.
${\displaystyle y-y_p=m(x-x_p)}$
y-(-7)=8(x-(-1))
y+7=8x+8
y=8x+1
Hence, the tangent line equation is: f(x)=8x+1

Explanation:
we require the slope m and a point (x,y) on the line
${\displaystyle m_{tangent}=f^{\prime }(-1)}$
${\displaystyle \Rightarrow f^{\prime }\left(x\right)=6-2x}$
${\displaystyle \Rightarrow f^{\prime }(-1)=6+2=8}$
and ${\displaystyle f(-1)=-6-1=-7\Rightarrow (-1,-7)}$
${\displaystyle \Rightarrow y+7=8(x+1)}$
${\displaystyle \Rightarrow y=8x+1\leftarrow }$ equation of tangent