Question

graph each equation.

1. \\(\dfrac{x^2}{4} + \dfrac{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 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 \(y\) - axis (major axis): Since \(a^{2}=9\), then \(a = 3\). The vertices are at \((0,\pm a)=(0, 3)\) and \((0,- 3)\).
• For the \(x\) - axis (minor axis): Since \(b^{2}=4\), then \(b = 2\). The co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points and draw the ellipse

Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane. Then, draw a smooth ellipse passing through these points. The ellipse will be centered at the origin \((0,0)\) because the standard form has no \(h\) or \(k\) (the center is \((h,k)=(0,0)\) for \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) when \(h = 0\) and \(k = 0\)).

Answer:

The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\), drawn by plotting the points \((0,3)\), \((0,-3)\), \((2,0)\), \((-2,0)\) and connecting them with a smooth curve. (The actual drawing would be an ellipse on the given coordinate grid passing through these points.)