Question

consider the following hyperbola.
\\( \frac { ( x + 3 ) ^ { 2 } } { 25 } - \frac { ( y + 5 ) ^ { 2 } } { 64 } = 1 \\)
step 1 of 3: find the coordinates of the vertices of the hyperbola.

Explanation:

Step1: Identify the standard form of hyperbola

The standard form of a hyperbola that opens left - right is \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\), where \((h,k)\) is the center of the hyperbola, and the vertices are \((h\pm a,k)\).
For the given hyperbola \(\frac{(x + 3)^2}{25}-\frac{(y + 5)^2}{64}=1\), we have \(h=-3,k = - 5,a^2 = 25\), so \(a = 5\).

Step2: Calculate the vertices

Using the formula for vertices \((h\pm a,k)\), substitute \(h=-3,k=-5,a = 5\).
For \(x\) - coordinate: \(x=-3\pm5\), \(y=-5\).
When \(x=-3 + 5=2\), the vertex is \((2,-5)\).
When \(x=-3-5=-8\), the vertex is \((-8,-5)\).

Answer:

The vertices of the hyperbola are \((-8,-5)\) and \((2,-5)\)