Find the equation of the tangent to the curve y = x2 -3x +5 at the point (1,3).
Find the equation of the tangent to the curve y = x2 -3x +5 at the point (1,3).The tangent is a straight line of the form y = mx+cTo find the gradient, m, we first differentiate the function:dy/dx = 2x -3Then we evaluate at the point (1,3). At this point x =1 , so dy/dx = 2(1) -3 = -1So m = -1, and therefore the tangent is of the form y = -x +cTo find c, we know that the point (1,3) must lie on the tangent and so we substitute these values for x and y into our equation:3 = -(1) +cRearranging gives c = 4Therefore the equation of the tangent is y = -x + 4
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