Find the eqn. of the tangent line to the curve of f(x) = x^2 + 5x -5 at (0,-5).
Differentiate f(x) to obtain an expression for the derivative (slope of the tangent line):
f '(x) = 2x + 5
Subst. 0 for x here: f '(0) = 2(0) + 5 = 5 (at the point (0, -5))
Use the point-slope equation of a str. line to find the eqn of the tan. line:
y-k = m(x-h), where (h,k) is a point on the line and m is the slope:
y - [-5] = 5(x-0), or y+5 = 5x. Then y = 5x - 5 is the eqn. of the TL to the given curve at (0,-5).
