To find the vertices of the hyperbola given by the equation
[tex]\[
\frac{x^2}{9} - \frac{y^2}{49} = 1,
\][/tex]
we start by recognizing the standard form of a hyperbola with a horizontal transverse axis. The standard form is:
[tex]\[
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.
\][/tex]
In our equation, we can identify [tex]\( a^2 \)[/tex] and [tex]\( b^2 \)[/tex]:
[tex]\[
a^2 = 9 \quad \text{and} \quad b^2 = 49.
\][/tex]
To find [tex]\( a \)[/tex] and [tex]\( b \)[/tex], we take the square roots of [tex]\( a^2 \)[/tex] and [tex]\( b^2 \)[/tex]:
[tex]\[
a = \sqrt{9} = 3 \quad \text{and} \quad b = \sqrt{49} = 7.
\][/tex]
The vertices of a hyperbola with a horizontal transverse axis are located at [tex]\( (\pm a, 0) \)[/tex]. Thus, for our hyperbola:
[tex]\[
\text{Vertices are at } (-a, 0) \text{ and } (a, 0).
\][/tex]
Substituting [tex]\( a = 3 \)[/tex]:
[tex]\[
\text{Vertices are at } (-3, 0) \text{ and } (3, 0).
\][/tex]
To follow the instruction to enter the smallest coordinate first, we order these coordinates correctly:
[tex]\[
((-3, 0), (3, 0)).
\][/tex]
Therefore, the vertices of the hyperbola are:
[tex]\[
((-3, 0), (3, 0)).
\][/tex]
