Find the equation of the tangent to the curve y = (5x+4)/(3x-8) at the point (2, -7).

Firstly, we need to find the gradient at this point on the curve as the tangent will have the same gradient. So to find the gradient of the curve or in other words, dy/dx, we differentiate the equation. To do this here, we use the quotient rule. I like to use a rhyme to help me remember the quotient rule - but it is also given in the formula sheet. I use "low d-high minus high d-low all over the square of what's below". Where low and high is the bottom and top of the fraction respectively and d-low and d-high is the differential of the bottom and top respectively.

So we get dy/dx = ((3x-8)*5 - (5x+4)*3)/(3x-8)², now we can substitute the x co-ordinate of the point given to find the gradient, which we get as -13. Now to find the equation of the tangent we can use y-y₁=m(x-x₁) where y1 and x1 are the cordinates of the point given. y--7=-13(x-2) simplifies to y+7=-13x+26 which in turn simplifies to y=-13x+19.