O.K.
(h; k) are the coordinates of the vertex.
On the graph are (3; 4), therefore we have:
[tex]f(x)=a(x-3)^2+4[/tex]
We have the x-intercept (1; 0) → x = 1; y = 0.
Substitute them in the equation:
[tex]0=a(1-3)^2+4\\\\0=a(-2)^2+4\\\\4a+4=0\ \ \ |-4\\\\4a=-4\ \ \ \ |:4\\\\a=-1[/tex]
so. We have the answer:
[tex]\boxed{f(x)=-(x-3)^2+4}[/tex]
