A curve C has equation y=(2x-3)^5. Find the equation of the normal of this curve at point P with y coordinate -32.

Finding normals and tangents to curves is a very common question in A-level maths papers, especially core 3 modules, giving between 5-8 marks depending on complexity. In order to start this question, the x coordinate of point p must be found. To do this, as we already know the y coordinate, we can simply substitute -32 in for y and solve for x. i.e -32=(2x-3)^5. To remove the power 5 on the right hand side, you will need to fifth root the left hand side (-32) which gives -2. Rearrange the equation to get the x terms on one side and numbers on the other; you should find the x coordinate to be 1/2. Now to find the equation of the normal, you need to find the gradient of the curve at the point P which we now know to have coordinates (1/2,-32). Differentiate the equation y=(2x-3)^5 by using the chain rule. By multiplying by the power and reducing the power by 1 and multiplying by the differential of the bracket, you should get dy/dx=10(2x-3)^4 Substitute your value of x into this new equation to find the gradient at the point P. You should find this to be 160. As we are looking for the normal, we use the negative reciprocal i.e gradient is -1/160. Using y-y1=m(x-x1) where m is the gradient, substitute the x and y values of point p into this equation. y--32=-1/160(x-1/2) You can multiply the left hand side by 160 to remove the fractional gradient of the right hand side to give: 160y+5120=-x+1/320 To tidy the equation up, I would suggest moving all the terms to the left hand side and equating to 0. 160y+x+5119.99=0

Unless specified, you can usually leave the answer either in the form ax+by+c=0 as above or y=mx+c.