Math question help please !!! urgent?

Question

whats the domain and range of

1/(x^2-4) and (1/x)^2-4

write an equation for the paraboloar with focus (0,2) and directrix Y=-2

find the focus and directrix for the parabola y=x^2-4x+4

Answer 1

whats the domain and range of

1/(x² - 4)

domain is when x² - 4 ≠ 0

x² ≠ 4

x ≠ ± 2

to find range:

y = 1/(x² - 4)

x² - 4 = 1/y

x² = (1 + 4y)/y

x = ± √((1 + 4y)/y)

so range:

y ≠ 0

and

(1 + 4y)/y > 0

When y > 0

1 + 4y > 0

4y > -1

y > -1/4 so y > 0

When y < 0

1 + 4y < 0

4y < -1

y < -1/4

So range is:

(-∞, -1/4) U (0, ∞)

(1/x)² - 4

domain is when

x ≠ 0

Range:

y = (1/x)² - 4

y + 4 = 1/x²

x² = 1/(y + 4)

x = ± √(1/(y + 4))

so y ≠ -4

and

1/ (y + 4) > 0

y + 4 > 0

y > -4

range is (-4, ∞)

write an equation for the parabola with focus (0,2) and directrix y = -2

half way between focus and directrix is the vertex, this is at (0, 0)

focal length is 2

Parabola is concave up

parabola has equation (x - h)² = 4a(y - k)

where (h, k) is the vertex and a is the focal length

x² = 8y is your parabola

find the focus and directrix for the parabola (concave up)

y = x² - 4x + 4

y = (x - 2)²

vertex is at (2, 0)

focal length:

4a = 1

a = 1/4

directrix

y = -1/4

focus = (2, 1/4)