Question

Minimum value of the polynomial $p\left(x\right)=4x^2-6x+1$

• A. $-\frac{3}{4}$
• B. $-\frac{5}{4}$
• C. $-\frac{5}{16}$
• D. $-\infty$

Answer 1

The correct option is B $-\frac{5}{4}$

$4x^2-6x+1$

Min value is at:- $\frac{-b}{2a}=x$

$\therefore \frac{6}{4\times 2}=x;x=\frac{3}{4}$

$\therefore P\left(\frac{3}{4}\right)=4{\left(\frac{3}{4}\right)}^2-6\left(\frac{3}{4}\right)+1$

$=4\frac{9}{16}-\frac{9}{2}+1$

$=\frac{9}{4}-\frac{9}{2}+1$

$=\frac{9-18+4}{4}=\frac{-5}{4}$