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Suppose a curve is defined by the function y = f(x) and you want the equation of the tangent at the point A.Suppose the x-coordinate at A is p. Then use y = f(p) to find the y-coordinate of A and let that be q. So, the point A is (p, q).
Next, find dy/dx, the derivative of the function f, with respect to q. This will be a constant or a functio of x. In the latter case, find its value for x = p. This is m, the gradient of the tangent.
So now you have the tangent line with gradient, m which passes through the point (p, q) and so its equation is
y - q = m(x - p)
Simplify this into the required form: y = mx + c or ax + by + c = 0.
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The answer will depend on the context. If the curve in question is a differentiable function then the gradient of the tangent is given by the derivative of the function. The gradient of the tangent at a given point can be evaluated by substituting the coordinate of the point and the equation of the tangent, though that point, is then given by the point-slope equation.
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