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What are the vertices of the hyperbola whose equation is (x+7)^2/25-(y-8)^2/32= 1?(−13, 8) and (−1, 8)
(−12, 8) and (−2, 8)
(−7, 2) and (−7, 14)
(−7, 3) and (−7, 13)

Respuesta :

Answer:

b, (-12,8) and (-2,8)

Step-by-step explanation:

graph it

Answer is B => (-12,8) and (-2,8)

Hyperbola :

The equation of the hyperbola is   [tex]\frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1[/tex]

then, the Center = (h, k).

And vertices = (h ± a, k)

=> (Given),

h = -7, k = 8

So, Center = (-7,8)

Now, [tex]a^{2} = 25[/tex]  => a = 5

[tex]b^{2} = 32[/tex]

i.e., Vertices of the hyperbola are = (-7 ± 5, 8)

=> (-7 + 5, 8) and (-7 - 5, 8)

=>  Vertices of hyperbola are (-2, 8) and (-12, 8).

Learn more about Hyperbola here:

https://brainly.com/question/16457232?referrer=searchResults

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