Respuesta :

parabola with vertx (3,-1) that opens down

4(-3)(y+1)=(x-3)^2
y+1=(-1/12)(x-3)^2
y=(-1/12)(x-3)^2-1

Answer:

The required quadratic function is [tex]y=-\frac{1}{12}(x-3)^2-1[/tex].

Step-by-step explanation:

The standard form of the parabola is

[tex](x-h)^2=4p(y-k)[/tex]

Where the focus is (h, k + p) and the directrix is y = k - p.

The directrix of y = 2 and a focus of (3, −4).

[tex](h,k+p)=(3,-4)[/tex]

On comparing both sides we get

[tex]h=3[/tex]

[tex]k+p=-4[/tex]                           ...... (1)

[tex]y=k-p[/tex]

[tex]k-p=2[/tex]                           ...... (2)

Add equation (1) and (2).

[tex]2k=-2[/tex]

[tex]k=-1[/tex]

Substitute k=-1 in equation (1).

[tex](-1)+p=-4[/tex]

[tex]p=-3[/tex]

Therefore the equation of parabola is

[tex](x-3)^2=4(-3)(y-(-1))[/tex]

[tex](x-3)^2=-12(y+1)[/tex]

It can be rewritten as

[tex]y=-\frac{1}{12}(x-3)^2-1[/tex]

Therefore the required quadratic function is [tex]y=-\frac{1}{12}(x-3)^2-1[/tex].

Ver imagen DelcieRiveria