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 + 1 or y = -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