Question

graph each equation.

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

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

Explanation:

Step1: Identify the ellipse standard form

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which matches the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (for a vertical major axis, since \(a^{2}>b^{2}\)). Here, \(a^{2}=9\) so \(a = 3\), and \(b^{2}=4\) so \(b = 2\). The center of the ellipse is at \((0,0)\) (the origin) because there are no shifts in \(x\) or \(y\) (the numerators are \(x^{2}\) and \(y^{2}\), not \((x - h)^{2}\) or \((y - k)^{2}\)).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis, since \(a\) is associated with \(y\)): The vertices are at \((0,\pm a)=(0,\pm3)\).
• For the \(x\) - axis (minor axis): The co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the center \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\).
• Plot the co - vertices \((2,0)\) and \((-2,0)\).

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be centered at the origin, with a vertical major axis (taller along the \(y\) - axis) passing through \((0,3)\) and \((0, - 3)\), and a horizontal minor axis passing through \((2,0)\) and \((-2,0)\).

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\). To draw it, plot these points and sketch a smooth curve connecting them.