If you fit the vertex and the y-intercept into that form of the parabola, what you get is this:
[tex]a(0-4) ^{2} =(15+1)[/tex] and
[tex]a(-4) ^{2} =16[/tex] and 16a=16. Solve that a to get that a = 1. Now you have your true parabola equation in vertex form, which is:
[tex]1(x-4) ^{2} =y+1[/tex] or just
[tex](x-4) ^{2} =y+1[/tex]
Follow my work here to find the x-intercepts:
[tex](x-4) ^{2} -1=y[/tex]
(x-4)(x-4)-1=y and [tex] x^{2} -8x+16-1=y[/tex]
Simplifying by combining like terms gives you this parabola:
[tex] x^{2} -8x+15=y[/tex]
In order to find the x-intercepts, or the roots of the parabola as they are also known as, you have to set y equal to 0 (because the x-intercepts are found when y = 0, right?)
[tex] x^{2} -8x+15=0[/tex]
Factoring that gives you (x-5)(x-3)=0 and that x-5=0 and x-3=0. Solving each of those simple equations gives you x=5 and x=3 when y=0. So the roots, in the form you require them, are (5, 0) and (3, 0)
