Hello,

to 1): The equation of a straight line is: y = mx + b. The slope m must be the same as the 1rt derivative of f if this line is a tangent to the graph of f.

Let T(t, 4t-tē) be the tangent point then

. Use point-slope-formula to calculate the equation of the tangent at T:

After a few steps of rearranging you'll get:

The given point is placed on this tangent too that means it's coordinate must satisfy this equation too. Plug in the coordinates:

and solve for t. I've got t= 1 or t = 3

Plug in these values of t into the equation of the tangent:

y = 2x + 1ory = -2x + 9

to 2): Given is . If a tangent to the graph of is parallel to y = x - 6 then the 1rt derivative equals 1:

. With x = -2 you get the equation:

After a few steps of rearranging you'll get k = 3.

Plug in this value into the equation of and you'll get