Question

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)

Answer 1

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)

Answer 2

Sent a picture of the solution to the problem (s). The problem with the last part was that you had a -2 instead of 2. The last equation is just slope. Sent all the steps in the pic with explanation.