Question

graph each equation.

1. $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$

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 \(b^{2} = 4\), so \(a = 3\), \(b=2\)) with a vertical major axis (because \(a^{2}\) is under the \(y^{2}\) term).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis) vertices: When \(x = 0\), from \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\), we get \(y^{2}=9\), so \(y=\pm3\). So the vertices are \((0, 3)\) and \((0,- 3)\).
• For the \(x\) - axis (minor axis) co - vertices: When \(y = 0\), from \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\), we get \(x^{2}=4\), so \(x=\pm2\). So the co - vertices are \((2,0)\) and \((- 2,0)\).

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 major axis is along the \(y\) - axis with length \(2a=6\) and the minor axis is along the \(x\) - axis with length \(2b = 4\).

Answer:

The graph is an ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) (plotted and connected as described above).