Question #cd906

1 Answer
Oct 25, 2017

Equation of tangent: #y=5x-2#, equation of normal: #y=-\frac{1}{5}x+\frac{16}{5}#

Explanation:

Tangents and normals are lines, which means to find their equations we need their gradients and a fixed point, which is #(1,3)# in this case.

We are given #y=3x^2-x+1#. Differentiating, we get #\frac{dy}{dx}=6x-1#. Thus the gradient of the tangent is given by #6(1)-1=5# and the gradient of the normal, perpendicular to the tangent, is given by #-\frac{1}{5}#. Thus the equation of the tangent is #y-3=5(x-1) \Leftrightarrow y=5x-2# and the equation of the normal is #y-3=-\frac{1}{5}(x-1) \Leftrightarrow y=-\frac{1}{5}x+\frac{16}{5}#.