Question 1024212
Convert to vertex form,
{{{f(x)=2x^2+8x-1}}}
{{{f(x)=2(x^2+4x)-1}}}
{{{f(x)=2(x^2+4x+4)-1-2(4)}}}
{{{f(x)=2(x+2)^2-9}}}
The vertex is at ({{{-2}}},{{{-9}}}) and the coefficient of the quadratic term is positive so the parabola opens upwards and the vertex value is a minimum.
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So the domain is all real numbers and the range is [{{{-9}}},{{{infinity}}}).
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*[illustration fc7.JPG].