Problem 16 - Intermediate Algebra

Question

The graph of each equation is a parabola. Find the vertex of the parabola and then graph it. $$y=-4(x-2)^{2}+2$$

Step-by-Step Solution

Answer

The vertex is (2, 2) and the parabola opens downward.

Step 1: Identify the Vertex Form

The function is in the vertex form of a parabola equation, which is written as \( y = a(x-h)^2 + k \) where \( (h, k) \) is the vertex of the parabola.

Step 2: Extract the Vertex

Based on the equation \( y = -4(x-2)^2 + 2 \), the vertex is \( (h, k) \). Here, \( h = 2 \) and \( k = 2 \). So the vertex is \( (2, 2) \).

Step 3: Identify Direction of Parabola

The coefficient \( a \) in the equation \( y = -4(x-2)^2 + 2 \) is \(-4\). Since \( a < 0 \), the parabola opens downward.

Step 4: Graph the Parabola

Plot the vertex at the point \((2, 2)\). Since the parabola opens downward and \( a \) is \(-4\), the graph will be narrower than a standard parabola that opens downwards. Sketch the parabola symmetrically around the vertical line \( x = 2 \).

Key Concepts

Vertex Form of a Parabola

When working with parabolas in algebra, the vertex form of the equation is extremely helpful. The vertex form is given by \( y = a(x-h)^2 + k \). Here, \((h, k)\) represents the vertex of the parabola, which is the turning point where the graph changes direction. This form allows you to easily identify the vertex and understand various properties of the parabola.
The vertex \((h, k)\) is very important because:

• It tells us the maximum or minimum value of the parabola.
• It is the point of symmetry of the parabola.

In the example equation \( y = -4(x-2)^2 + 2 \), the vertex \((h, k)\) can be directly identified as \((2, 2)\). Recognizing the vertex form simplifies the task of identifying these critical elements without the need for further calculations.

Parabola Graphing

When graphing a parabola, knowing the vertex form \( y = a(x-h)^2 + k \) can be incredibly useful. Once you've identified the vertex, you can plot it as a starting point.
In the example \( y = -4(x-2)^2 + 2 \), the vertex is plotted at \((2, 2)\).
From the vertex, the parabola will either ascend or descend depending on the value of \( a \).

• If \( a > 0 \), the parabola opens upwards, rising from the vertex.
• If \( a < 0 \), it opens downwards, falling from the vertex.

After plotting the vertex and determining the direction, you can sketch the parabola. Make sure it is symmetric around the vertical line \( x = h \). In this case, the line is \( x = 2 \).
Use additional points by choosing values for \( x \) left and right of the vertex to get more accuracy.

Direction of a Parabola

The direction in which a parabola opens is determined by the coefficient \( a \) in its vertex form \( y = a(x-h)^2 + k \). Understanding this direction is crucial for predicting the behavior of the parabola.
- If \( a > 0 \), the parabola opens upwards, looking like a U-shape.- If \( a < 0 \), as in our example \( a = -4 \), the parabola opens downwards like an upside-down U.
The magnitude of \( a \) affects how "wide" or "narrow" the parabola appears:

• Larger absolute values of \( a \) (e.g., 4 or -4) make the parabola narrower.
• Smaller absolute values of \( a \) (e.g., 0.5 or -0.5) make it wider.

In \( y = -4(x-2)^2 + 2 \), \( a = -4 \) indicates that the parabola opens downward and is narrower compared to a parabola with \( a = -1 \). Understanding the sign and magnitude of \( a \) helps in accurately sketching the parabola's shape.