The curve C has equation y=(2x-3)^5, the point P lies on C and has coordinates (w, – 32), find (a) the value of w and (b) the equation of the tangent to C at the point P in the form y=mx+c , where m and c are constants.

(a) The curve is defined by y=(2x-3)^5. To find x=w when y=-32, we must substitute these values into the equation C and re-arrange to find w. -32=(2w-3)^5. First we must remove the power of 5 by doing power of 1/5 to each side, so (-32)^1/5=(2w-3). (-32)^1/5=-2 so -2=2w-3. Then we add 3 to both sides so 2w=1. Finally we divide both sides by 2 so w=1/2.(b) The equation of any tangent at the point (x1,y1) is given by y-y1=m(x-x1) where m is the value of the gradient at that point. First, we need to work out the value of the gradient by differentiating our original C equation with respect to x. We do this by using the chain rule: y=(2x-3)^5 so dy/dx = 5d(2x-3)/dx(2x-3)^4 so dy/dx = 52(2x-3)^4 = 10(2x-3)^4. To find the gradient, we substitute our value of x1 (aka 1/2) into our equation: dy/dx = 10(2(1/2)-3)^4 = 160. Now we have our value of m we can work out the equation of the tangent at he point P: y+32=160(x-1/2), now we re-arrange to find in the form of y=mx+c: y+32 = 160x -80, y = 160x -112