In this problem, we have a vertical parabola open downward
The vertex is a maximum
Remember that
The distance from the vertex to the directrix must be the same that the distance from the vertex to the focus
so
step 1
Find out the distance from the directrix to the focus
directrix -------> y=3
Focus ----> (1,1/2)
distance=3-1/2=2.5
2.5/2=1.25
the vertex is the midpoint between the focus and the directrix
so
the coordinates of the vertex are (1,0.5+1.25)
vertex (1,1.75)--------> vertex (1.7/4)
The equation in vertex form is given by
[tex]y=a(x-h)^2+k[/tex]
where
a is the leading coefficient
(h,k) is the vertex
(h,k)=(1,7/4)
substitute
[tex]y=a(x-1)^2+\frac{7}{4}[/tex][tex]y=a(x^2-2x+1)+\frac{7}{4}[/tex]
Remember that
[tex]-(x-h)^2=4p(y-k)[/tex]
where
p is the focal distance
In this problem
p=1.25=5/4
substitute (h,k)=(1,7/4)
[tex]-(x-1)^2=4(\frac{5}{4})(y-\frac{7}{4})[/tex][tex]-x^2+2x-1=5y-\frac{35}{4}[/tex]
Isolate the variable y
[tex]\begin{gathered} 5y=-x^2+2x-1+\frac{35}{4} \\ 5y=-x^2+2x+\frac{31}{4} \\ \\ y=-\frac{1}{5}x^2+\frac{2}{5}x+\frac{31}{20} \end{gathered}[/tex]
The answer is option C
