Question 749467
Checking how the vertex and the y intercept are arranged on a graph, we can understand that the parabola opens upward, having the vertex as a minimum.


{{{y=a(x-h)^2+k}}} standard form in which {{{a>0}}}.
Using your vertex coordinates,
{{{y=a(x-4)^2-1}}}


The given y intercept point allows the equation, {{{15=a(0-4)^2-1}}}
{{{15=a*16-1}}}
{{{16=a*16}}}
{{{a=1}}}


Standard form equation is then specifically {{{y=1*(x-4)^2-1}}}
or simply
{{{highlight(y=(x-4)^2-1)}}}


From that standard form equation, you can finish finding the x intercepts on your own.  Just let y = 0 and solve for x.  {{{(x-4)^2-1=0}}}.