Consider the function and the value of a. f(x) = 5 - x - x^2, a = 0
(a) Use Mtan = lim h->0 [f(a + h) – f(a)] / h
to find the slope of the tangent line mtan = f'(a).

Mtan=1-2a

Step-by-step explanation:

Mtan=lim h->0[5-(a+h)-(a+h)^2-(5-a-a^2)/h

=lim h->0[5-a-h-a^2-2ah-h^2-5+a+a^2]/h

=lim h->0[h-2ah-h^2+5-5-a+a-a^2+a^2]/h

=lim h->[h-2ah-h^2 +0+0+0]/h

=lim h->o[h(1-2a-h)/h

=lim h->0 (1-2a-h)

=1-2a-0

=1-2a

Mtan=1-2a.

true !

step-by-step explanation:

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)