# 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.

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