Question

directrix of y=2 and focus of (3,-4) what quadratic function is created?

f(x)=1/12(x-3)^2-1
f(x)=1/6(x+3)^2+1
f(x)=1/6(x-3)^2+1
f(x)=-1/12(x-3)^2-1

Answer 1

@robtobey @SolomonZelman

Answer 2

As the directrix is y = 2 and the focus is at (3,-4) then the vertex of this parabola is midway between these features at (3, -1).

The parabola is a downward opening parabola.

The general form of such a parabola centered at the origin is x² = 4cy where c is negative and represents the distance between the vertex and the focus.

When the vertex is not at the origin but at (e,f) then the general equation becomes (x -

e)² = 4c(y - f).

In this question (e, f) = (3, -1) and c = -3

Then the equation becomes (x - 3)² = 4*-3(y + 1) = -12(y + 1) = -12y - 12

After re-arranging with y as the subject : 12y = -12 - (x - 3)² thus y = f(x) = -1 - (x - 3)²/ 12

Answer: f(x) = -1/12 (x - 3)^2 - 1

Answer 3

Thank you!

Answer 4

You're welcome!! Xo!