Question

Minimum value for the polynomial ${4x}^2-6x+1$ is:

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

Answer 1

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

Given,

$4x^2-6x+1$

The formula for finding the minimum value is at $-\frac{b}{2a}=x$

$\therefore -\frac{(-6)}{2\times 4}=x$

$\Rightarrow x=\frac{3}{4}$

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

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

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

$=-\frac{5}{4}$