Question

-

Answer 1

\(\bf 4 = 2^2 \qquad \qquad 25 = 5^2\\\\ \quad \\
\cfrac{(x+2)^2}{4}-\cfrac{(y+4)^2}{25}= 1 \implies \cfrac{(x+2)^2}{2^2}-\cfrac{(y+4)^2}{5^2}= 1\\
\cfrac{(x+2)^2}{2^2}-\cfrac{(y+4)^2}{5^2}= 1 \implies \cfrac{(x-h)^2}{a^2}-\cfrac{(y-k)^2}{b^2}=1\)

so, where do you think is the center of that hyperbola?

notice the minus between the fractions, thus is a hyperbola
notice that the POSITIVE fraction is the one with the "x" variable, thus is a "horizontal traverse axis"

Answer 2

center is (-2, 4)?

Answer 3

(-2 , -4 )

from there, the center, you move
"a" units over the "x-axis"
and "b" units over the "y-axis"

is a horizontal traverse axis hyperbola, so the vertex will be at the ( h + a , k) and ( h - a, k)

and the covertex will be at ( h, k + b) and ( h, k - b)

Answer 4

So center is (-2, -4), how do i then get the vertices?

Answer 5

see above ^

Answer 6

so (2, -4) and (2, 21)?

Answer 7

did i do that correctly?

Answer 8

?

Answer 9

notice, your a = 2, b = 5
h = -2 , k = -4

Answer 10

the vertices will be at (-2+2, -4) and (-2 -2, -4)
covertices will be at (-2, -4+5) and (-2, -4-5)