Work out the equation of the tangent to the curve y=x^2+5x-8 at the point (2,6)

Firstly, we know that dy/dx is the gradient of the tangent to a curve. If we know the gradient of the tangent, and the point on the curve where the tangent touches it, we can work out the equation of the tangent.Now, we will find dy/dx. If y=xn, then we know that dy/dx=nxn-1So if y= x2+5x-8, then dy/dx=2x+5.At the point (2,6), where x=2, the gradient of the tangent is dy/dx= 2(2)+5=9. The general equation of a straight line is y=mx+c, where m is the gradient.So, to find the equation of the tangent, we will sub in m=9. So y=9x+c.At the point (2,6), x=2 and y=6, which implies 6=9(2) +c, so c=-12.Therefore, the equation of the tangent is y=9x-12.

AD
Answered by Alexandra D. Further Mathematics tutor

6738 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

write showing all working the following algebraic expression as a single fraction in its simplest form: 4-[(x+3)/ ((x^2 +5x +6)/(x-2))]


What is the range of solutions for the inequality 2(3x+1) > 3-4x?


How can you divide an algebraic expression by another algebraic expression?


Differentiate y = x*cos(2x)


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