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
Answered by Jack G. Maths tutor

3310 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I draw and sketch an equation?


Find 1 + (1 + (1 + (1 + (1 + ...)^-1)^-1)^-1)^-1


How do I integrate ln(x)?


Find the turning points of the curve y=2x^3 - 3x^2 - 14.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences