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

Related Maths A Level answers

All answers ▸

Find dy/dx when x+2y+3y^2= 2x^2+1


how do integrate an equation with a surd or a fraction?


Find the general solution of 2 dy/dx - 5y = 10x


What are the conditions for an event to be modeled with the binomial distribution?