The normal to the curve \(y = 2x^{2} + x - 3\) at the point (2, 7) meets the x- axis at the point P. Find the coordinates of P.
\(y = 2x^{2} + x - 3\)
\(\frac{\mathrm d y}{\mathrm d x} = 4x + 1\)
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).
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