Find the equation of the straight line tangent to the curve y=2x^3+3x^2-4x+7, at the point x=-2.

We are looking for a straight line, so it needs the form y=mx+c. To find our gradient, m, we need the gradient of the curve at the point x=-2, so differentiate the equation: dy/dx=6x2+6x-4, and solve at x=-2, ie m=64-62-4=8.To find c, calculate the y coordinate at x=-2 using the equation of the curve: y=2*(-8)+34-4(-2)+7=11. Using the values we have for y, m and x, we can calculate what value c should be: y=mx+c, so c=y-mx=11-8*(-2)=27.Thus the equation for the tangent line at x=-2 is y=8x+27.

Answered by James B. Maths tutor

5615 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A curve C has equation y = x^2 − 2x − 24sqrt x, x > 0. Prove that it has a stationary point at x=4.


Calculate dy/dx of the following equation: y = 3x^3 - 6x^2 + 2x - 6


Differentiate: y = 4x^3 - 5/x^2


Find dy/dx such that y=(e^x)(3x+1)^2.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences