The normal to the curve \(y = 2x^{2} + x - 3\) at the point

Explanation

\(y = 2x^{2} + x - 3\)

\(\frac{\mathrm d y}{\mathrm d x} = 4x + 1\)

At (2, 7), the gradient is \(4(2) + 1 = 9\)

The gradient of normal = \(-1 \div 9 = \frac{-1}{9}\)

Equation of the normal = \(y - 7 = \frac{-1}{9}(x - 2)\)

\(9y - 63 = 2 - x\)

At the point of meeting the x- axis, y = 0

\(0 - 63 = 2 - x \implies x = 65\)

The coordinate of P = (65, 0).