Find the equation of the tangent to the curve y = 4x^2 (x+3)^5 at the point (-1, 128).

y = 4x2(x+3)5 . Use the product rule to find the first derivative of the curve, 8x(x+3)5 + 20x2(x+3)4 , and substitute x = -1 to find the gradient at the point (-1, 128). This should be 64. Now substitute x = -1 and y = 128 into the equation y = mx + c where m = 64 and c is the unknown y-intercept. Solving the equation shows that c = 192. The equation of the tangent is y = 64x + 192.

JG

Related Maths A Level answers

All answers ▸

Can you explain the product rule when differentiating?


Find the range of a degree-2 polynomial function such as 2x^2 +1, or x^2 + 2x - 3.


Differentiate: y = sin(2x).


Solve the equation 2cos2(x) + 3sin(x) = 3, where 0<x<=π