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

Answered by Caleb R. Maths tutor

4720 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I find the stationary points on the curve y = f(x) = x^3+6x^2-36x?


How would you differentiate 3x^4 - 2x^2 + 9x - 1


A curve has an equation of y = 20x - x^2 - 2x^3, with one stationary point at P=-2. Find the other stationary point, find the d^2y/dx^2 to determine if point P is a maximum or minium.


What does dy/dx represent?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences