Question

graph each equation.

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

coordinate plane with x from -8 to 8 and y 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 \(b^2=4\), so \(a = 3\), \(b = 2\)) with a vertical major axis (because \(a>b\) and the \(y\)-term has the larger denominator).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis) vertices: When \(x = 0\), we solve for \(y\). Substitute \(x = 0\) into the equation \(\frac{0^2}{4}+\frac{y^2}{9}=1\), which simplifies to \(\frac{y^2}{9}=1\), so \(y^2=9\) and \(y=\pm3\). So the vertices are \((0, 3)\) and \((0,- 3)\).
• For the \(x\) - axis (minor axis) co - vertices: When \(y = 0\), we solve for \(x\). Substitute \(y = 0\) into the equation \(\frac{x^2}{4}+\frac{0^2}{9}=1\), which simplifies to \(\frac{x^2}{4}=1\), so \(x^2 = 4\) and \(x=\pm2\). So the co - vertices are \((2,0)\) and \((- 2,0)\).

Step3: Plot the points

• Plot the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\) on the coordinate plane.
• Then, sketch the ellipse by connecting these points smoothly, making sure the ellipse is symmetric about both the \(x\) - axis and \(y\) - axis.

To graph the ellipse:

1. Mark the points \((0,3)\), \((0, - 3)\), \((2,0)\), and \((-2,0)\) on the given coordinate grid.
2. Draw a smooth curve connecting these points, forming an ellipse that is wider along the \(y\) - axis (since the major axis is vertical with length \(2a = 6\)) and narrower 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. (The actual graph is an ellipse centered at the origin \((0,0)\), with a vertical major axis of length 6 and a horizontal minor axis of length 4.)