WILL MARK BRAINLIEST PLEASE HELP

Using 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

Respuesta :

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]







Ver imagen kudzordzifrancis

Answer:

For flvs students the answer is

f(x) = −1/8(x − 4)^2 + 3

Step-by-step explanation: