Problem 41 - College Algebra

Question

Find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$(y+1)^{2}=-8 x$$

Step-by-Step Solution

Answer

The vertex is at (0, -1), the focus is at (-2, -1), and the line x = 2 is the directrix. The parabola faces leftwards.

Step 1: Identify the Vertex (h, k)

Rewrite the given equation \((y+1)^2=-8x\) in the standard form. The vertex (h, k) will be \(-(h)\) in \(y+1\) and \(-k\) in \(x-h\). In this case, \(k=-1\) and \(h=0\), so the vertex is \((0, -1)\).

Step 2: Determine the Focus and Directrix

Next we find the value of 4p from the equation. The coefficient of \(x\) in the equation is \(-8\), so we equate \(-8 = -4p\). Solving for \(p\) yields \(p = 2\). Remember that for left-facing parabolas the focus is \(h-p, k\) and the directrix line equation is \(x=h+p\). Substituting \(h=0\) and \(p=2\) into the formulas gives us the focus \((-2, -1)\) and the directrix \(x = 2\).

Step 3: Graph the Parabola

First, plot the vertex (0, -1). It's the highest or lowest point in the graph. Then plot the focus \((-2, -1)\), just to the left of the vertex, showing that the parabola faces leftwards. Draw the line \(x=2\) to represent the directrix. Now you can sketch the graph of parabola, which will be a U shape opening to the left towards the focus and away from the directrix.