Question 345419
{{{f(x)=-(1/2)x^2+2x+3}}}
Complete the square to put the function into vertex form, {{{y=a(x-h)^2+k}}} where ({{{h}}},{{{k}}}) is the vertex.
{{{f(x)=-(1/2)(x^2-4x)+3}}}
{{{f(x)=-(1/2)(x^2-4x+4)+3+4/2}}}
{{{f(x)=-(1/2)(x-2)^2+5}}}
a) {{{h=2}}}
b) {{{k=5}}}
c) The vertex lies on the axis of symmetry, {{{x=2}}}
d) {{{k=5}}} since the coefficient of {{{x^2}}} is negative the parabola opens downward and the value at the vertex is a maximum.
{{{ymax=5}}}
e) {{{h=2}}}
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{{{drawing(300,300,-8,8,-8,8,grid(1),circle(2,5,0.2),blue(line(2,-10,2,10)),graph(300,300,-8,8,-8,8,-(1/2)x^2+2x+3))}}}