WILL MARK BRAINLIEST PLEASE HELPUsing a directrix of y = 5 and a focus of (4, 1), what quadratic function is created? f(x) = 1/4(x − 4)^2 − 3 f(x) = 1/8(x + 4)^2 − 3 f(x) = −1/8(x − 4)^2 + 3 f(x) = −1/4(x + 4)^2 − 3

Answer:
[tex]f(x)=-\frac{1}{8}(x-4)^2+3[/tex]Step-by-step explanation:
The directrix given to us has equation,
[tex]y=5[/tex] and the focus is [tex](4,1)[/tex].  This means that the axis of symmetry of the parabola is parallel to the y-axis and has equation [tex]x=4[/tex], because it must go through the focus.
This axis of symmetry of the parabola will meet the directrix at [tex](4,5)[/tex] .
The vertex of this parabola is the midpoint of the point of intersection of the axis of symmetry and the focus.
Thus,
[tex]V(h,k)=(\frac{4+4}{2},\frac{5+1}{2})[/tex]
[tex]V(h,k)=(4,3)[/tex].
The equation is given by [tex](x-h)^2=4p(y-k)[/tex].
[tex](x-4)^2=4p(y-3)[/tex].
[tex]|p|[/tex] is the distance between the vertex and the focus, which is 2.
This implies that,
[tex]p=-2[/tex] or [tex]p=2[/tex]
But the position of the directrix and the vertex implies that  the parabola opens downwards.[tex]\therefore p=-2[/tex].
The equation of the parabola now becomes;
[tex](x-4)^2=4(-2)(y-3)[/tex].

[tex](x-4)^2=-8(y-3)[/tex]
We solve for y to obtain,
[tex]y=-\frac{1}{8}(x-4)^2+3[/tex]
or
[tex]f(x)=-\frac{1}{8}(x-4)^2+3[/tex]