Write the equation of a hyperbola with a center at (-5, -3), vertices at (-5, -5) and (-5, -1) and co-vertices at (-11, -3) and (1, -3).

The equation is (y + 3)²/4 - (x + 5)²/36 = 1

Step-by-step explanation:

* Lets revise the equation of the hyperbola

- The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the y-axis is

  (y - k)²/a² - (x - h)²/b² = 1

# The length of the transverse axis is 2a

# The coordinates of the vertices are (h , k ± a)

# The length of the conjugate axis is 2b

# The coordinates of the co-vertices are (h ± b , k)

# The distance between the foci is 2c, where c² = a² + b²

# The coordinates of the foci are (h , k ± c)

* Lets solve the problem

∵ The center of the hyperbola is (-5 , -3)

∵ The coordinates of the its center is (h , k)

∴ h = -5 and k = -3

∵ Its vertices are (-5 , -5) and (-5 , -1)

∵ The coordinates of its vertices are (h , k + a) and (h , k - a)

∴ k + a = -5 and k - a = -1

∵ k = -3

∴ -3 + a = -5 and -3 - a = -1

∵ -3 + a = -5 ⇒ add 3 to both sides

∴ a = -2

∵ Its co-vertices are (-11 , -3) and (1 , -3)

∵ The coordinates of the co-vertices are (h + b , k) and (h - b , k)

∵ h = -5

∴ -5 + b = -11 and -5 - b = 1

∵ -5 + b = -11 ⇒ add 5 to both sides

∴ b = -6

∵ The equation of it is (y - k)²/a² - (x - h)²/b² = 1

∵ h = -5 , k = -3 , a = -2 , b = -6

∴ The equation is (y - -3)²/(-2)² - (x - -5)²/(-6)² = 1

∴ The equation is (y + 3)²/4 - (x + 5)²/36 = 1

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

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