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]