Question 994328
{{{f(x)=-(1/2)x^2-3x+5/2}}}
{{{f(x)=-(1/2)(x^2+6x)+5/2}}}
{{{f(x)=-(1/2)(x^2+6x+9)+5/2+(1/2)9}}}
{{{f(x)=-(1/2)(x+3)^2+7}}}
So the vertex is ({{{-3}}},{{{7}}}) and the axis of symmetry is {{{x=-3}}}.
To get the y-intercept, set {{{x=0}}}.
{{{y=5/2}}}
({{{0}}},{{{5/2}}})
To get the x-intercepts, set {{{y=0}}}.
{{{0=-(1/2)(x+3)^2+7}}}
{{{(1/2)(x+3)^2=7}}}
{{{(x+3)^2=14}}}
{{{x+3=0 +- sqrt(14)}}}
{{{x=-3 +- sqrt(14)}}}
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*[illustration cr1.JPG].