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

4610 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

I don't understand the point of differentiation or integration


∫ (ln(x)/(x*(1+ln(x))^2) dx


The sum of the first K natural numbers is 300. Find the value of K.


Integrate y=2x^2 +4x-1


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