What is the equation of the quadratic graph with a focus of (1, 1) and a directrix of y = −1?

f(x) = − one fourth (x − 1)2 + 1

f(x) = − one fourth (x − 1)2

f(x) = one fourth (x − 1)2 + 1

f(x) = one fourth (x − 1)2

Alright,
directix is y=something so it opens down or up

we use
(x-h)²=4p(y-k)
the vertex is (h,k)
and p is distance from focus to vertex
if focus is above directix, p is positive
if focus is below directix, p is negative

so we gts

focus=(1,1)
directix is y=-1
1>-1
focus is above

oh, vertex is in middle of focus and directix
so
beteeen (1,1) and y=-1 is, hmm
that is a distance of 2 vertically
2/2=1
1 down from (1,1) is (1,0)
vertex is (1,0)
p=1

so
(x-1)²=4(1)(y-0)
solving for y to get into f(x)=something form
(x-1)²=4y
y=1/4(x-1)²
f(x)=1/4(x-1)²

4th option

The 3rd quadrant as 3x+g5ti=56i: 92d-543