Question

passing through (6,2) and tangent to the line x-4y-15=0 at (3,-3)

Answer 1

-first of all make `y` the subject of the formula
-then differentiate the curve to get `dy/dx`
-substitute `x value` of the meeting point of the curve with your tangent (in this case `x=3`, in your `dy/dx`, this is the gradient of your tangent
- recall `y=mx+c` where mis the gradient
-since your line passes through the point `(6,2)` substitute x=6 and y=2
-solve for c
now your equation is complete