Your post, "F(x)=x^2+5x-5 points (0,f(0)) equation of the tangent line," is a challenge to read. Did you mean the following?
F(x)=x^2+5x-5. Find the equation of the tangent line at (0, f(0))
If so, differentiate F(x) to obtain an expression for the slope of the tangent line:
F '(x) = 2x + 5
Now let x=0, to find the slope of the tangent line at (0,f(0)):
F '(0) = 2(0) + 5 = 5
Now you have the slope of the tangent line to the curve at (0, [0^2+5(0) - 5]).
Find the equation of the line with this slope that passes thru (0, -5):
y - [-5] = 5(x-[-5]), or y+5 = 5(x+5), or y = -5 + 5x + 25.
Final answer: y = 5x + 20.
