Final answer:
The vertices of the ellipse 2x² + 9y² = 81 are located at (4.5, 0) and (-4.5, 0).
Step-by-step explanation:
To find the vertices of the ellipse 2x² + 9y² = 81, we first need to rewrite the equation in the standard form of an ellipse, which is (x²/a²) + (y²/b²) = 1, where 'a' and 'b' are the semi-major and semi-minor axes, respectively.
Dividing the entire equation by 81 gives us x²/40.5 + y²/9 = 1. This simplifies to x²/(4.5²) + y²/(3²) = 1, where a is 4.5, and b is 3.
The vertices of an ellipse are found at ±a along the x-axis and ±b along the y-axis. For this equation, the vertices along the x-axis are at (4.5, 0) and (-4.5, 0), and since b is less than a, there are no vertices along the y-axis (as those would be co-vertices).