Question 1199204


{{{f(x)=x^2+x-2}}}

put the equation in vertex form: {{{f(x)=(x-h)^2+k}}}

{{{f(x)=x^2+x-2}}}...complete the square

{{{f(x)=(x^2+x+b^2)-b^2-2}}}......{{{b=1/2}}}

{{{f(x)=(x^2+x+(1/2)^2)-(1/2)^2-2}}}

{{{f(x)=(x+1/2)^2-1/4-2}}}

{{{f(x)=(x+1/2)^2-9/4}}}



determine the vertex point
=>{{{h=-1/2}}},{{{k=-9/4}}}

the vertex point is:({{{-1/2}}},{{{-9/4}}})

 the {{{x}}}-intercept  : set {{{f(x)=0}}}

{{{0=(x+1/2)^2-9/4}}}

{{{(x+1/2)^2=9/4}}}

{{{sqrt((x+1/2)^2)=sqrt(9/4)}}}

{{{x+1/2=3/2}}}=>{{{x=3/2-1/2}}}=>{{{x=1}}}

or

{{{-(x+1/2)=3/2}}}=>{{{-x-1/2=3/2}}}=>{{{-x=3/2+1/2}}=>{{{-x=2}}=>{{{x=-2}}



the {{{y}}}-intercept : set {{{x=0}}}

{{{f(0)=(0+1/2)^2-9/4}}}
{{{f(0)=1/4-9/4}}}
{{{f(0)=8/4}}}
{{{f(0)=2}}}

and the axis of symmetry is:{{{x=-1/2}}}


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(-1/2,-9/4,.12),locate(-1/2,-9/4,V(-1/2,-9/4)),
green(line(-1/2,10,-1/2,-10)),
graph( 600, 600, -10, 10, -10, 10,(x+1/2)^2-9/4)) }}}