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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…

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 \(b^{2}=4\), 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): 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 vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\).
  • Then, sketch the ellipse by connecting these points smoothly, making sure that the ellipse is symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

The graph is an ellipse with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2, 0)\), \((-2, 0)\), symmetric about the \(x\) - axis and \(y\) - axis, and centered at the origin \((0,0)\). (To draw it, plot the four points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and sketch the ellipse through them.)