Point A lies on the curve y=3x^2+5x+2. The x-coordinate of A is 2. Find the equation of the tangent to the curve at the point A

First differentiate the function with respect to x, dy/dx=6x+5 this finds the gradient function now calculate the gradient at point A by inputing x=2 into the gradient function 6(2)+5=17. Now using y=mx+c where m is known gives y=17x+c now must solve for c, at x=2 y=24 by 3(2)^2+5(2)+2=24 now we can solve for c where 24=17(2)+c this gives c=-10 y=17x-10

DS
Answered by Dylan S. Further Mathematics tutor

4391 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

A curve has equation y = ax^2 + 3x, when x= -1, the gradient of the curve is -5. Work out the value of a.


How to solve the inequality 1 - 2(x - 3) > 4x


Point A lies on the curve: y=x^2+5*x+8. The x-coordinate of A is -4. What is the equation of the normal to the curve at A?


Factorise 6x^2 + 7x + 2


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