Question

graph each equation.

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

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 \(a > b\) for vertical major axis), where \(a^2 = 9\) and \(b^2=4\). So \(a = 3\) and \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis, since \(a\) is under \(y^2\)): 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 draw an ellipse passing through these four points, centered at the origin \((0,0)\) (since the equation has no \(x\) or \(y\) shifts, i.e., \(h = 0\), \(k = 0\) in the general ellipse equation \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\)).

Answer:

The graph is an ellipse centered at the origin with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((- 2,0)\), drawn through these points. (To actually graph it, plot the four points and sketch the ellipse connecting them.)