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
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