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 the gradient of y^2 +2xln(y) = x^2 at the point (1,1)


What does differentiating do?


Consider the curve y=x/(x+4)^0.5. (i) Show that the derivative of the curve is given by dy/dx= (x+8)/2(x+4)^3/2 and (ii) hence find the coordinates of the intersection between the left vertical asymptote and the line tangent to the curve at the origin.


A 1kg mass is launched from the ground into the air at an angle of 30 degrees to the horizontal and with initial speed 25 ms^-1. Assuming negligible air resistance, how far from the starting point will the mass travel before it hits the ground?