Question

Find the min value of $x^2-3x+6$

• A. $\frac{-15}{4}$
• B. $\frac{15}{4}$
• C. $0$
• D. None.

Answer 1

Answer: (A)

$\frac{-15}{4}$

Given expression $x^2-3x+6$

The expression has least point at $f^{\prime }\left(x\right)=0⟹2x-3=0x=\frac{3}{2}$

So the min values is ${\left(\frac{3}{2}\right)}^2-3\left(\frac{3}{2}\right)+6\frac{9}{4}-\frac{9}{2}+6-\frac{9}{4}+6=-\frac{15}{4}$