Questions: What is (are) the value(s) of x at the vertex (vertices) of the shape defined by the equation x^2/64 + y^2/36 = 1? (Separate multiple values of x with a comma.)

Solution Steps

The given equation is $\frac{x^2}{64} + \frac{y^2}{36} = 1$. This is the standard form of an ellipse centered at the origin $(0, 0)$ with semi-major and semi-minor axes.

In the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the values $a^2 = 64$ and $b^2 = 36$ are given. The semi-major axis is along the x-axis because $a^2 > b^2$.

The length of the semi-major axis is given by $a = \sqrt{64} = 8$.

The vertices of the ellipse along the x-axis are located at $(\pm a, 0)$. Therefore, the x-coordinates of the vertices are $x = -8$ and $x = 8$.

Final Answer