Question 1106926
So, the first derivative is,
{{{dy/dx=2ax+b}}}
and the second is,
{{{d2y/dx2=2a}}}
So,
{{{2a=2}}}
{{{a=1}}}
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{{{y=x^2+bx+c}}}
Converting to vertex form,
{{{y=(x^2+bx+(b/2)^2)+(c-(b/2)^2)}}}
{{{y=(x+b/2)^2+(c-(b/2)^2)}}}
So the minimum occurs at,
{{{-b/2=-1/2}}}
{{{b=1}}}
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{{{y=x^2+x+c}}}
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Finally using the point,
{{{-4=(-2)^2+(-2)+c}}}
{{{-4=4-2+c}}}
{{{c=-6}}}
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{{{highlight(y=x^2+x-6)}}}