Consider the following (slope and functions!)

Question

Function	Line
f(x) = −1/16x2	x + y = 0

(a) Find f'(x).

f'(x)= ?

(b) Find the point on f'(x) where the tangent line is parallel to the given line.

(x, y) = (____)

(c) Find an equation of the line tangent to the graph of f'(x) and parallel to the given line.

y= ?

 

 

Answer 1

x+y=0 is a negative sloping line that crosses the origin and has a slope of -1x because x+y=0->x=-y. If the tangent line is parallel to this line then it must also have a slope of -1. First we find the derivative of this function -1/16x^2->-1/8x then since the line has a slope of -1 we can assume that the value for the coordinates for x and y are the same like (1,-1),(2,-2),(3,-3) etc. so plug in one of these points to y=-1/8x so y=-1/8(2)-------------->y=-1/4 then use point-slope form to find the equation. (y-(-1/4)=-1(x-2)->y+1/4=-1x+2->y=-1x+7/4 so the equation for the tangent line is y=-1x+7/4 and the point were the tangent line is paralell to the given line is (2,-1/4).