Ihave the problem solved but i dont know how they got it. an equation for the line tangent to y= -5-9x^2 at (2,-41). i use limh→0 f( x0+h)-f(x0)/h to get the slope of tan line which is -36, then it says i need to get the slope-intercept form with the equation below and i'm not sure how -36= )/)

Hello : 
f(x) = 5-9x²
x0 = 2  f(2) = 5-9(2)² = - 31
f(2+h) = 5 - 9(2+h)² = -9h²- 36h -31 
subsct in :  limh→0 f( x0+h)-f(x0)/h
 limh→0 (- 9h²-36h -31 -(-31))/h =  limh→0 ( -9h² - 36h ) /h
 =  limh→0 h(-9h -36)/h
simplify by : h 
 limh→0 (-9h - 36) = - 36.. ( slope of the tangent at :  (2,-41).
an equation is : y - ( -41) = -36 ( x - (-2))
so : 
-36= y-(-41)/x-(-2)