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| The slope of the normal to the curve \(y=x^{2}+3x+2\) at the point \((-2,0)\) is |
\(-1\) \(1\) \(\frac{1}{2}\) \(-\frac{1}{2}\) |
| \(1\) |
| \(\begin{aligned}\left(\frac{dy}{dx}\right)_{(-2,0)}&=2x+3\left|_{(-2,0)}\right.\\ &=-1\end{aligned}\hspace{6 cm}\) Thus Slope of normal \(=1\) |
