Jambmaths question:
Find the minimum value of $y={{x}^{2}}-2x-3$
Option A:
1
Option B:
–1
Option C:
–4
Option D:
4
Jamb Maths Solution:
$\begin{align} & y={{x}^{2}}-2x-3 \\ & \frac{dy}{dx}=2x-2 \\ & \text{At stationary point }\frac{dy}{dx}=0 \\ & 2x-2=0 \\ & x=1 \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=2>0 \\ & \text{At }x=1\text{ is a minimum point} \\ & \text{The value of }y\text{ at }x=1 \\ & y={{1}^{2}}-2(1)-3=-4 \\\end{align}$
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