Question 1045147
You make better sense of this when put h into standard form.


{{{y=a(x-h)^2+k}}}, standard form.


{{{y=a(x-4)^2+7}}} and you have a point,  (2,0).
Think of how a parabola is symmetric.
Can you judge that {{{a<0}}} ?


{{{a(x-h)^2+7=y}}}
{{{a(x-h)^2=y-7}}}
{{{a=(y-7)/(x-h)^2}}}
{{{a=(0-7)/(2-4)^2}}}
{{{a=-49/2}}}


{{{y=-(49/2)(x-4)^2+7}}}
Find BOTH zeros!