Question 948400
 H(x)=3x-1/2x^2 Write the equation in the form:
{{{ax^2+bx+c=0}}}
In this case:
{{{(-1/2)x^2+3x=y}}}
 The x-intercept is where y=0 so
{{{(-1/2)x^2+3x+0=0}}}*[invoke quadratic "x", -1/2, 3, 0 ]
The answers are 0 and 6, so the x-intercepts are (0,0) and (6,0)
And the y-intercept(s) is where x=0, so:
{{{y=(-1/2)0^2+3(0)}}}
y=0 So the y-intercept is at the origin (0,0) 
The axis of symmetry is given by: (This is the x value of the vertex)
{{{x=-b/2a}}}
In this case:
{{{x=(-3)/(2(-1/2))}}}
x=3 Substitute this in the original equation to get the y value of the vertex.
{{{y=(-1/2)3^2+3(3)}}}
{{{y=(-1/2)(9)+9}}}
{{{y=-4.5+9}}}
y=4.5 So the vertex is (3,4.5)