Question

graph each equation.

1. \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\)

graph with x-axis from -8 to 8, y-axis from -8 to 8, grid lines

Explanation:

Step1: Identify the conic section

The equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (since \(a^{2}=9\) and \(b^{2} = 4\), and \(a>b\), so the major axis is along the \(y\)-axis).

Step2: Find the vertices and co - vertices

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices (end - points of the major axis) are \((0,\pm a)\) and the co - vertices (end - points of the minor axis) are \((\pm b,0)\).
Given \(a^{2}=9\), then \(a = 3\) (we take the positive value since it represents a length), so the vertices are \((0, 3)\) and \((0,-3)\).
Given \(b^{2}=4\), then \(b = 2\) (we take the positive value since it represents a length), so the co - vertices are \((2,0)\) and \((- 2,0)\).

Step3: Plot the points

• Plot the vertices \((0,3)\) and \((0, - 3)\) on the \(y\) - axis.
• Plot the co - vertices \((2,0)\) and \((-2,0)\) on the \(x\) - axis.

Step4: Draw the ellipse

Connect the plotted points with a smooth curve to form the ellipse. The ellipse will be centered at the origin \((0,0)\) (since the standard form of the ellipse has no \(h\) or \(k\) terms, where the center is \((h,k)\)) and will be wider along the \(y\) - axis (because \(a>b\)).

Answer:

The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\), drawn by connecting the points \((0,3)\), \((0, - 3)\), \((2,0)\), and \((-2,0)\) with a smooth curve. (The actual graph is an ellipse on the given coordinate grid with the described key points.)