Question

The graph point of the polynomial p(x) = $2x^2+4x+3$ has its least value at the point

• A. $(-1,9)$
• B. $(1,9)$
• C. $(-1,1)$
• D. $(0,3)$

Answer 1

The correct option is C $(-1,1)$

$\text{f}\left(\text{x}\right)\text{=2}{\text{x}}^{\text{2}}\text{+4x+3}$
$\text{f}{}^{\prime }\left(\text{x}\right)\text{=4x+4}$
$\text{for}\text{critical}\text{values}\text{set}\text{f}{}^{\prime }\left(\text{x}\right)\text{=0}$
$\text{f}{}^{\prime }\left(\text{x}\right)\text{=0}$
$\text{4x+4=0}$
$\text{x=-1}$
$\text{f}{}^{\prime }{}^{\prime }\left(\text{x}\right)\text{=4}$
$\text{at}\text{x=-1}\text{f}{}^{\prime }{}^{\prime }\left(\text{x}\right)\text{is}\text{positive}$
$\text{hence}\text{minimum}\text{at}\text{x=-1}$
$\text{y=2}{\left(\text{-1}\right)}^{\text{2}}\text{+4}\left(\text{-1}\right)\text{+3}$
$\text{=2-4+3}$
$\text{=1}$
$\text{Hence}\text{minimum}\text{point}\text{is}\left(\text{-1,1}\right)$