Question 182219
</pre><font size=4><b>
Given {{{system((1/2)x^2 + x -1=0)}}}
{{{f(x)=ax^2+bx+c}}}, where a, b, and c are real numbers, with {{{a<>0}}}
Remember:{{{system(a=1/2,b=1,c=-1)}}}

To find x-coordinate of the vertex--->{{{x=-b/(2a)}}}
{{{x=-1/(2*(1/2))=-1/(cross(2)*1/cross(2))}}}
{{{highlight(x=-1)}}}
To find y-Intercept:
{{{y=(1/2)(-1)^2+(-1)-1=(1/2)-2=(1-4)/2=-3/2}}}
{{{highlight(y=-3/2)}}}
Vertex---->(-1,-3/2)
The Axis of symmettry is {{{x=-1}}}.
Solving for X-Intercepts:
By Quadratic formula: 
{{{x=(-b+-sqrt(b^2-4ac))/(2a)=(-1+-sqrt(1^2-4(1/2)(-1))))/(2(1/2))}}}
{{{x=(-1+-sqrt(1+2))/1=(-1+-sqrt(3))/1}}}
{{{x=(-1+1.732)/1=highlight(0.732)}}}
{{{x=(-1-1.732)/1=highlight(-2.732)}}}
To find Y-Intercept:
{{{f(x)=0}}}
{{{y=(1/2)(0)^2+0-1=highlight(-1)}}}
We'll mark all highlighted points in the graph:
{{{drawing(400,400,-5,5,-5,5,grid(1),graph(400,400,-5,5,-5,5),blue(circle(-1,-3/2,.12)),circle(0.732,0,.12),circle(-2.732,0,.12),circle(0,-1,.12),green(line(-1,5,-1,-5)))}}}---->{{{drawing(400,400,-5,5,-5,5,grid(1),graph(400,400,-5,5,-5,5,(1/2)x^2+x-1),blue(circle(-1,-3/2,.12)),circle(0.732,0,.12),circle(-2.732,0,.12),circle(0,-1,.12))}}}
Thank you,
Jojo</pre>