Question 936451
{{{y=a(x-4)^2-1}}}
So when {{{x=0}}},
{{{15=a(0-4)^2-1}}}
{{{15=a(16)-1}}}
{{{16a=16}}}
{{{a=1}}}
So,
{{{y=(x-4)^2-1}}}
{{{(x-4)^2=1}}}
{{{x-4=0 +- 1}}}
{{{x=4 +- 1}}}
{{{x=3}}} and {{{x=5}}}