Question 948333
f(x)=2x^2+5x+3
Section A: Y-intercept
The y-intercept is the y value(s) where x=0
In this case:
{{{y=2(0^2)+5(0)+3}}}
{{{y=3}}} ANSWER A The y-intercept is (0,3)
Section B: X-intercept
The graph intercepts the x axis when the y value is 0, so:
{{{2x^2+5x+3=0}}}
*[invoke quadratic "x", 2, 5, 3 ]
Thus the x-intercepts are at -1 and -1.5 or
ANSWER B: X=intercepts are at (-1,0) and (-1.5,0).
Section C: Vertex
Given equation of form:
{{{f(x)=Ax^2+Bx+C}}} 
the axis of symmetry (x value of vertex) is given by:
{{{x=((-B)/2A)}}}
In this case:
{{{x=((-5)/2(2))}}}={{{-5/4}}}
Use this x value in the equation to find y value of vertex
{{{y=2(-5/4)^2+5(-5/4)+3}}}
{{{y=2(25/16)-(25/4)+3}}}
{{{y=(25/8)-(25/4)+3}}}
{{{y=-25/8+24/8}}}={{{-1/8}}} ANSWER C: The vertex is at (-5/4,-1/8) 
This graph shows the vertex more clearly, with x- and y- intercepts:
{{{ graph( 800, 800, -5, 5, -5, 5, 2x^2+5x+3) }}}