Question

Question #cd906

Answer 1

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}`.