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 ▸

Integrate Sin(x)Cos(x)dx.


What are the parameters of the Poisson distribution?


Determine the stationary points of y=(5x^2)/(lnx)


If z1 = 3+2i, z2= 4-i, z3=1+i, find and simplify the following: a) z1 + z2, b) z2 x z3, c)z2* (complex conugate of z2), d) z2/z3.