Create a line that is tangent to f(x)=3-5x^2 and goes through point (-1,-2).

The given point (-1, -2) lies on the graph of the given function:  f(-1) = 3 - 5 = -2.
In this situation, the equation of the tangent line at this point is
y = f'(-1)(x - (-1)) + f(-1).
Because  f'(x) = -10x,  we obtain  f'(-1) = 10  and the equation becomes
y = 10(x + 1) - 2 = 10x + 8.
The graph is attached.
 
[the math editor is broken, says "f(x)=x^2" for many formulas]

We are going to make a tangent line of the form
eq. (1): y(x)=m*x +b
Such that y(x) goes through point (-1,-2). Therefore y(-1)=-2 must be satisfied.
Here m is the slope that is tangent at point (-1,-2) and b is the y-intercept.
First lets find m. 
We need to find the derivative f'(x) which will give the slope of any point x on the line f(x). Hence, m=f'(-1) since that is the point we are interested in.
f'(x)=(3-5x^2)'=-10x
m=f'(-1)=-10(-1)=10
Now plug m into eq. (1). Solve for b such that x and y go through the point of interest.
y(x)=mx +b
-2=10(-1)+b
8=b
You now have your line that is tangent line at the point (-1,-2).
y(x)=10x +8