I have 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 = y - (-41) / x - (-2)

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)