Question

sketch a graph of $(x + 3)^{2}-\frac{(y - 3)^{2}}{4}=1$

Explanation:

Step1: Identify the standard form of hyperbola

The standard form of a hyperbola is \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) (opens left - right).
For the given equation \((x + 3)^2-\frac{(y - 3)^2}{4}=1\), we have \(h=-3,k = 3,a = 1,b = 2\).

Step2: Find the center

The center of the hyperbola is \((h,k)=(-3,3)\).

Step3: Find the vertices

The vertices of a hyperbola of the form \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) are \((h\pm a,k)\).
Substituting \(h=-3,a = 1,k = 3\), the vertices are \((-3 + 1,3)=(-2,3)\) and \((-3-1,3)=(-4,3)\).

Step4: Find the asymptotes

The equations of the asymptotes of a hyperbola \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) are \(y - k=\pm\frac{b}{a}(x - h)\).
Substituting \(h=-3,k = 3,a = 1,b = 2\), the asymptotes are \(y-3=\pm2(x + 3)\).

• For \(y-3 = 2(x + 3)\), we get \(y=2x+6 + 3=2x + 9\).
• For \(y-3=-2(x + 3)\), we get \(y=-2x-6 + 3=-2x - 3\).

To sketch the graph:

1. Plot the center at the point \((-3,3)\).
2. Plot the vertices at \((-2,3)\) and \((-4,3)\).
3. Draw the asymptotes \(y = 2x+9\) and \(y=-2x - 3\) as dashed lines.
4. Sketch the hyperbola approaching the asymptotes.

Answer:

Sketch a hyperbola with center \((-3,3)\), vertices \((-2,3)\) and \((-4,3)\), and asymptotes \(y = 2x+9\) and \(y=-2x - 3\) on the given coordinate - plane.