graph each equation.\n9) \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\)\ngraph with x-axis from -8 to 8 and y-axis…

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 - intercepts (vertices, since the major axis is vertical), set $x = 0$. Then $\frac{0^2}{4}+\frac{y^2}{9}=1\Rightarrow y^2=9\Rightarrow y=\pm3$. So the vertices are $(0, 3)$ and $(0, - 3)$.
• For the x - intercepts (co - vertices), set $y = 0$. Then $\frac{x^2}{4}+\frac{0^2}{9}=1\Rightarrow x^2 = 4\Rightarrow x=\pm2$. 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 draw a smooth ellipse passing through these points. The ellipse will be centered at the origin $(0,0)$ (since there are no shifts in the $x$ or $y$ terms in the equation $\frac{x^2}{4}+\frac{y^2}{9}=1$), with a vertical major axis of length $2a=6$ (from $y=- 3$ to $y = 3$) and a horizontal minor axis of length $2b = 4$ (from $x=-2$ to $x = 2$).

Answer:

The graph is an ellipse centered at the origin with vertices at ((0, \pm3)) and co - vertices at ((\pm2, 0)), drawn by plotting these points and connecting them with a smooth curve. (The actual drawing would be on the given coordinate grid with the points ((0,3)), ((0, - 3)), ((2,0)), ((-2,0)) plotted and a smooth ellipse through them.)