Let's break down the equation and understand the components step-by-step to find the center and vertices of the hyperbola described by the equation [tex]\(\frac{(y-4)^2}{6^2}-\frac{(x+9)^2}{7^2}=1\)[/tex].
### 1. Identify the Center of the Hyperbola
The general form for a vertically oriented hyperbola is:
[tex]\[
\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1
\][/tex]
In this equation:
- The center [tex]\((h, k)\)[/tex] is derived from the variables in the squared terms.
For the given equation [tex]\(\frac{(y-4)^2}{6^2} - \frac{(x+9)^2}{7^2} = 1\)[/tex]:
- The term [tex]\((y - 4)^2\)[/tex] indicates [tex]\(k = 4\)[/tex].
- The term [tex]\((x + 9)^2\)[/tex] indicates [tex]\(h = -9\)[/tex].
So, the center of the hyperbola is:
[tex]\[
\boxed{(-9, 4)}
\][/tex]
### 2. Identify the Vertices
For a vertically oriented hyperbola centered at [tex]\((h, k)\)[/tex], the vertices are at points [tex]\((h, k \pm a)\)[/tex]. Here:
- [tex]\(a = 6\)[/tex], which is the value below the term involving [tex]\(y\)[/tex].
For the given center [tex]\((-9, 4)\)[/tex]:
- The bottom vertex is at [tex]\((h, k - a) = (-9, 4 - 6) = (-9, -2)\)[/tex].
- The top vertex is at [tex]\((h, k + a) = (-9, 4 + 6) = (-9, 10)\)[/tex].
So the vertices are:
[tex]\[
\boxed{(-9, -2)}
\][/tex]
and
[tex]\[
\boxed{(-9, 10)}
\][/tex]
In summary:
- The center of the hyperbola is [tex]\((-9, 4)\)[/tex].
- The bottom vertex of this vertically oriented hyperbola is [tex]\((-9, -2)\)[/tex].
- The other vertex is [tex]\((-9, 10)\)[/tex].
You can now fill in the boxes accordingly:
- The center of the hyperbola is [tex]\(\boxed{(-9, 4)}\)[/tex].
- The left vertex (bottom vertex in this vertical case) is [tex]\(\boxed{(-9, -2)}\)[/tex].
- The other vertex is [tex]\(\boxed{(-9, 10)}\)[/tex].
