Find the tangent for the line y=x^3+3x^2+4x+2 at x=2

Firstly, differentiate the equation y=x³+3x²+4x+2 to find the gradient function. The gradient function is dy/dx and is found to be 3x²+6x+4 = f'(x) after differentiation. The simple rule is to multiply the term by the power of x, and then subtract one from the power to differentiate each term. Also you should remember to remove the constant when differentiating (in this case "2") because the power of x is 0 and multiplying a term by 0 gives 0. Then, to find the gradient at the point x=2, simply put the number 2 into the gradient function to find the gradient. So, the gradient m=f'(2) = 28. To find the tangent, first find the y coordinate where x=2 by substituting this into the original equation (the first equation). This is found to be 30. So the coordinates are (2,30) of where the tangent will cross. Now, apply the equation y-y1=m(x-x1), where (x1,y1) are the coordinates of the tangent. The equation looks like this: y-30=2(x-28) This turns out to be, after rearranging, y=2x-26.