In part B, how would I factor the point-slop equation? Because I do not quite understand how the answer was -36x+40, for part b. Transcribed

In part B, how would I factor the point-slop equation? Because I do not quite understand how the answer was -36x+40, for part b. Transcribed Image Text: Thus, the slope of the curve at P(2, – 32) is – 36.
b. Use this slope and the given point (2, – 32) to write the equation for the tangent line.
y -(- 32) = – 36(x- (2))
This is the point-slope equation.
y =
(Simplify your answer. Do not factor.) Transcribed Image Text: a. Start with a secant line through P(2, – 32) and Q(2 + h,4 – 9(2 + h)2) nearby. Write an expression for the slop
the secant PQ, and determine what happens to the slope as Q approaches P; that is, as h approaches 0.
Ду
Secant slope
Ax
(4 – 9(2 + h)²) – (4 – 9(2)²)
%3D
h
Simplify the numerator.
36h –
h – 9n?
Ду
Secant slope
Дх
(Simplify your answer. Do not factor.)
h
Now simplify the entire fraction to get – 36 – 9h. value does this expression approach as h approaches 0
– 36
Thus, the slope of the curve at P(2, – 32) is – 36.