Problem 9 - Algebra for College Students

Question

For Problems \(1-30\), find the vertex, focus, and directrix of the given parabola and sketch its graph. $$ x^{2}=12(y+1) $$

Step-by-Step Solution

Answer

Vertex: (0, -1); Focus: (0, 2); Directrix: y = -4.

Step 1: Identify Parabola Form

The given equation of the parabola is in the form \(x^2 = 4p(y-k)\), where \((h, k)\) is the vertex. Comparing \(x^2 = 12(y+1)\) with \(x^2 = 4p(y-k)\), we identify \(4p = 12\) and \(k = -1\).

Step 2: Calculate p

From \(4p = 12\), solving for \(p\) gives \(p = \frac{12}{4} = 3\).

Step 3: Find the Vertex

The vertex \((h, k)\) is \((0, -1)\). This comes from rearranging the given equation; the vertex form indicates \(h = 0\) and \(k = -1\).

Step 4: Determine the Focus

The focus of a parabola with vertical orientation is located at \((h, k+p)\). Thus, the focus is at \((0, -1 + 3) = (0, 2)\).

Step 5: Find the Directrix

The directrix of the parabola is the line \(y = k - p\). Therefore, the directrix is \(y = -1 - 3 = -4\).

Step 6: Sketch the Parabola

Plot the vertex at \((0, -1)\), focus at \((0, 2)\), and draw the directrix line at \(y = -4\). Sketch the parabola opening upwards, as it is symmetric around the y-axis and opens in the positive y-direction.

Key Concepts

Vertex

The vertex of a parabola is a crucial point that often serves as a guide for sketching its graph. It is where the parabola changes direction. In mathematical terms, the vertex is the coordinate \(h, k\). For the given equation

• \(x^2 = 12(y+1)\),
• the vertex is found by rearranging the equation to match the standard form \(x^2 = 4p(y - k)\).

Here, we identify the vertex as \(0, -1\).
The vertex tells us where the parabola reaches its minimum or maximum value. For this vertical parabola, the vertex at \(0, -1\) is the lowest point, meaning the parabola opens upwards. This means that the y-values increase as we move away from the vertex along the parabola.

Focus

The focus of a parabola is a special point inside the curve where all the reflected lines off the parabola converge. When you're dealing with a parabolic shape, the focus \(F\) can help understand the direction and openness of the parabola.

• For a vertically oriented parabola like our example \(x^2 = 12(y+1)\),
• the focus is located at \(h, k + p\).

Since we've established \(h = 0\), \(k = -1\), and \(p = 3\),
the focus is at \(0, 2\).
This tells us more about the parabola since points inside reflect back to this focus. It also helps in sketching the parabola, providing a point to ensure the curve is drawn with correct symmetry and orientation.

Directrix

The directrix is a line that, together with the focus, helps determine the shape and location of a parabola. It lies opposite the focus relative to the vertex and below the parabola when it's oriented upwards.

• For our parabola \(x^2 = 12(y+1)\),
• the directrix is a horizontal line defined by \(y = k - p\).

Using \(k = -1\) and \(p = 3\), we find the directrix line at \(y = -4\).
Understanding its role:

• The directrix contributes to maintaining the shape of the parabola by defining a set distance from any point on the parabola to both the directrix and the focus.
• Every point on the parabola is equidistant from the focus and the directrix.

Quadratic Equations

Quadratic equations are foundational in understanding parabolas. They generally take the form \(ax^2 + bx + c = 0\), but for parabolas, we often see a rearrangement to make interpretation easier. For vertical parabolas, you might see them in the form \(x^2 = 4p(y - k)\), which closely relates to our example.

• These equations represent parabolas due to the quadratic (squared) term, dictating their characteristic U-shape.
• Learning to rearrange quadratic equations to match standard forms like \(x^2 = 4p(y - k)\) simplifies locating key components like the vertex, focus, and directrix.

Understanding quadratic equations helps predict and sketch paths of parabolas in both real-life contexts and graphically.