Question 145531
You have equations for parabolas.


The vertex of the general parabola {{{f(x)=ax^2+bx+c}}} is at the point (x,y) where {{{x=-(b/2a)}}} and {{{y=f(-(b/2a))}}}.  You just need to calculate {{{-(b/2a)}}} and then evaluate the function at that value.


The axis of symmetry is the vertical line that passes through the vertex, and the equation is {{{x=-(b/2a)}}}.  You can get this from the first step above.


The y-intercept is where the graph crosses the y-axis, that is to say, where {{{x=0}}}.  Therefore the y-intercept is the point ({{{0}}},{{{f(0)}}}).  You just need to evaluate {{{f(0)}}}.


The x-intercepts are at the values of x where the function equals 0.  So set your function equal to zero and solve the resulting quadratic equation by whatever means necessary (actually, they factor, and both of your equations will have integer roots).  The two solutions you get will be the x-coordinates of the x-intercepts, and the y-coordinates will be 0.