Actually, it is quite simple to do. For instance, let's define the equation symbolically in MATLAB:
syms y(x)
y(x)=4*x-3*x^2;
Then, take the derivative:
dy(x)=diff(y,x)
which would be
4-6*x
At this point, you can find the slope of the tangent line at point (2,-4) by inserting 2 into the above equation, which would be
4-6*(2)=-8
You know that the slope of tangent line is -8, but you should also find the value of y for that tangent line. Consider it as
y=-8*x+c
Then find the c by simply
-4=-8*(2)+c
where c would be 12.
If you plot both curves in the same figure, you will get
where the blue line is the original equation and red line is tangent line.
