Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y plus five squared divided by nine = 1

Answer:The vertices are (3 , -5) , (-5 , -5)The foci are (4 , -5) , (-6 , -5)Step-by-step explanation:* Lets study the equation of the hyperbola- The standard form of the equation of a hyperbola with     center (h , k) and transverse axis parallel to the x-axis is   (x - h)²/a² - (y - k)²/b² = 1- The length of the transverse axis is 2 a - The coordinates of the vertices are  (h  ±  a  ,  k)- The coordinates of the foci are (h ± c , k), where c² = a² + b²- The distance between the foci is  2c* Now lets solve the problem- The equation of the hyperbola is (x + 1)²/16 - (y + 5)²/9 = 1* From the equation# a² = 16 ⇒ a = ± 4# b² = 9 ⇒ b = ± 3# h = -1# k = -5∵ The vertices are (h + a , k) , (h - a , k)∴ The vertices are (-1 + 4 , -5) , (-1 - 4 , -5)* The vertices are (3 , -5) , (-5 , -5)∵ c² = a² + b²∴ c² = 16 + 9 = 25∴ c = ± 5∵ The foci are (h ± c , k)∴ The foci are (-1 + 5 , -5) , (-1 - 5 , -5)* The foci are (4 , -5) , (-6 , -5)