Sure, we can find all these for the given equation that is in the form of a hyperbola. The equation of the hyperbola ((Y+3)^(2))/(9)-((X+1)^(2))/(4)=1 is given in its standard form.
1. To find the center of the hyperbola, we look at the values inside the brackets along with the Y and X terms in the equation. The center (h, k) in general hyperbola equation is shifted from the origin (0,0) using subtractive transformation (The form is (Y-k) and (X-h) ). Hence in our given equation, compared to general form, k is -3 (because we have (Y+3) ) and h is -1 (because we have (X+1) ). So, the center of our hyperbola is (-1, -3).
2. The vertices of the hyperbola are given by (h, k ± a). From the equation we can see that "a" is the square root of denominator term associated with Y, which is 3 (because the denominator term is 9 that is 'a' square in general form). So, plugging in our values of h, k and a, we get the vertices as (-1, -3-3) and (-1, -3+3), which simplifies to (-1, -6) and (-1, 0).
3. The foci are given by (h, k ± sqrt(a^2 + b^2)). Here, "b" is the square root of the denominator term associated with X, which is 2 (because the denominator term is 4 that is 'b' square in the general form). Using these, a^2 + b^2 becomes 9 + 4 = 13. So, plugging in our values of h, k, a, and b, we get the foci as (-1, -3- sqrt(13)) and (-1, -3+ sqrt(13)). These numerically equate to approximately (-1, -6.60555127546399) and (-1, 0.6055512754639891).
So, the center of the hyperbola is (-1, -3), the vertices are at (-1, -6) and (-1, 0), and the foci are approximately at (-1, -6.60555127546399) and (-1, 0.6055512754639891).