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, origin at (0,0)

Explanation:

Step1: Identify the ellipse type

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which 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 \(y\)-intercepts (vertices, since major axis is vertical), set \(x = 0\):

\(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\Rightarrow y^{2}=9\Rightarrow y=\pm3\). So the vertices are \((0, 3)\) and \((0, - 3)\).

• For the \(x\)-intercepts (co - vertices), set \(y = 0\):

\(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\Rightarrow x^{2}=4\Rightarrow x=\pm2\). 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, sketch a smooth ellipse passing through these points. The ellipse will be centered at the origin \((0,0)\), stretched 3 units up and down along the \(y\)-axis, and 2 units left and right along the \(x\)-axis.

Answer:

The graph is an ellipse centered at \((0,0)\) with vertices at \((0, \pm3)\) and co - vertices at \((\pm2, 0)\), and it is drawn by plotting these points and sketching a smooth curve through them. (The actual graphing involves marking the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) on the given coordinate grid and connecting them with a smooth elliptical curve.)