Question

graph the hyperbola given by the following equation\\(\frac{(x + 5)^2}{4} - \frac{y^2}{12} = 1\\)

Explanation:

Step1: Identify Hyperbola Type

The equation is \(\frac{(x + 5)^2}{4}-\frac{y^2}{12}=1\), which is a horizontal hyperbola (since the \(x\)-term is positive) in the form \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\), where \((h,k)=(-5,0)\), \(a^2 = 4\) (\(a = 2\)), \(b^2=12\) (\(b = 2\sqrt{3}\)).

Step2: Find Vertices

Vertices are at \((h\pm a,k)\), so \((-5\pm2,0)\), i.e., \((-3,0)\) and \((-7,0)\). The graph shows vertices (the black dots) aligned horizontally, matching a horizontal hyperbola shifted left by 5 units.

Step3: Check Asymptotes and Shape

Asymptotes for horizontal hyperbola: \(y - k=\pm\frac{b}{a}(x - h)\), so \(y=\pm\sqrt{3}(x + 5)\). The graph's branches open left and right, consistent with horizontal hyperbola. The center at \((-5,0)\) (midpoint of vertices) and the \(a = 2\), \(b = 2\sqrt{3}\) match the graph's scale (distance from center to vertices is 2, and the "opening" related to \(b\)).

Answer:

The given graph correctly represents the hyperbola \(\frac{(x + 5)^2}{4}-\frac{y^2}{12}=1\) (the graph shows a horizontal hyperbola with center \((-5,0)\), vertices at \((-7,0)\) and \((-3,0)\), and appropriate branch directions).