Find the equation of the tangent to the curve y = x^2-2x-3 at x=-1

First we find the gradient by differentiation. Differentiating the expression for the curve gives dy/dx=2x-2. Subbing in x=-1 gives dy/dx=-4 so the gradient of the line is -4.
To find the y intercept, we use the formula for a straight line: y = mx+c. Rearranging we obtain y-mx=c. We then find y at the point x=-1 by subbing this into our original expression for the curve and get y=0. m is the gradient we have just obtained (m=-4). So we find c=0-(-4)*(-1)=-4 and so the expression for the tangent line at x=-1 is y=-4x-4

CR
Answered by Caleb R. Maths tutor

5703 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

What is a derivative?


Find dy/dx at t=3, where x=t^3-5t^2+5t and y=2t^2


Given that y=(4x-3)^3 x sin2x find dy/dx


How do I determine the domain and range of a composite function, fg(x) ?


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