Find the equation of the tangent to the curve y = 3x^2 + 4 at x = 2 in the form y = mx + c

There are two main steps. First find the gradient of the curve at x = 2 (m). This is done by differentiating the curve equation y = 3x^2 + 4 to get dy/dx = 6x. By plugging in x = 2, we get the gradient of the tangent, m, as 62 = 12. Then we need to find the y intercept of the tangent, c. We make c the subject, so c = y - mx. We worked out what m is (12) so we just need a set of coordinates x,y which lie on the tangent. The easiest point is where the tangent meets the curve. We know x = 2 so plugging that into the curve equation gives y = 3(2^2) + 4 = 16. Now we have values of x,y,m we can find c. c = 16 - 12*2 = -8. Therefore the final answer is y = 12x - 8

Answered by Maths tutor

5514 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A circle with equation x^2+y^2-2x+8y-40=0. Find the circle centre and the radius


You are given that n is a positive integer. By expressing (x^2n)-1 as a product of factors, prove that (2^2n)-1 is divisible by 3.


The curve C has equation y = x^3 - 3x^2 - 9x + 14. Find the co-ordinates and nature of each of the stationery points of C.


The function f(x) is defined by f(x) = 1 + 2 sin (3x), − π/ 6 ≤ x ≤ π/ 6 . You are given that this function has an inverse, f^ −1 (x). Find f^ −1 (x) and its domain


We're here to help

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

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning