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

6765 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Lengths of two sides of the triangle and the angle between them are known. Find the length of the third side and the area of the triangle.


Why is it that when 'transformation A' is followed by 'transformation B', that the combined transformation is BA and not AB?


A=(1,a;0,1/2) B=(1,-1;0,2) AB=I, calculate the value of a.


Find the General Second Order Differential Equation Using Substitution (A2 Further Maths)


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