The curve C has the equation y=((x^2+4)(x-3))/2*x where x is not equal to 0 . Find the tangent to the curve C at the point where x=-1 in the form y=mx+c

Firstly we need to expand out y into a series of terms to make it easier for us to compute the derivative . You multiply out the brackets to get y=(x³-3x²+4x-12)/2x then we divide each of the terms of the numerator by 2x , y=1/2x²-3/2x+2-6x^-1 (where 6x^-1=6/x). Now we differentiate y in term of x . In order to do this we multiply the term by the current power and subtract 1 from the current power i.e. y=x^n. dy/dx=nx^(n-1), dy/dx= x - 3/2 +6x^-2. . To find the gradient of C we must substitute the x value (-1) into the derivative of y . When a line is a tangent it has the same gradient as the original line so the tangent will have the same gradient as C , dy/dx= -1-3/2+6 = 7/2 . We are not given the y coordinate so we must substitute the x coordinate into the orginal equation to give us the y value at x=-1 , y=((1+4)(-1-3)) / -2 = 10. So we can use the general equation of a line , y=mx+c where m= gradient and x and y are points on the line , to find the constant c . 10=7/2(-1) +c We now rerange this equation to find the value of the constant c , c=10+7/2=27/2 . We now have all the pieces need to form the equation , so we substitute c back into the equation of a line , y=7/2x +27/2. This is now the equation of the line in the form y=mx+c