Question 40

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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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