Question 83362
for a parabola of the form y=ax^2+bx+c, the equation of the axis of symmetry is x=-b/2a ... in this case x=5/2


the vertex lies on the axis of symmetry and can be found by substituting the x value from the axis of symmetry into the equation


y=(5/2)^2-5(5/2)+3=(25/4)-(25/2)+3=-13/4 ... vertex is ((5/2),(-13/4))


a good table of values would be ... (-1,9), (0,3), (1,-1), (2,-3), (3,-3), (4,-1), (5,3), (6,9)


since the coefficient of the x^2 term is positive, the parabola opens upward


the y-intercept is in the table of values (0,3)


there are 2 x-intercepts, one between (0,3) and (1,-1); and another between (4,-1) and (5,3)