Find the equation of the normal to the curve y = 3x 2 +2 at

Explanation

y = \(3x^2 + 2\)

\(y^1 = \frac{dy}{dx} = 6x\)

Evaluating this derivative at x = 1 (since the point of interest is (1, 5)) gives us the slope of the tangent line at that point:

\(^mtangent = y^1(1) = 6 (1) = 6\)

Slope of the Normal Line \( ^mnorma l= - \frac{1}{^mtangent}\)

\(^mnormal = - \frac{1}{6}\)

\(y−y_1= ^mnormal⋅(x−x_1)\)

=y-5=-\(\frac{1}{6}(x-1)\)

=y-5=-\(\frac{1}{6}x+\frac{1}{6}\)

=y=-\(\frac{1}{6}x+\frac{1}{6}+5\)

Multiply through by 6

=6y=-x+1+30

∴6y+ x - 31=0