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