Problem 15 - Algebra 2

Question

Graph each function. $$ y=\frac{1}{4}(x-2)^{2}+4 $$

Step-by-Step Solution

Answer

The graph is a parabola with vertex (2,4) opening upwards, wider than the standard \( y=x^2 \).

Step 1: Identify the Function Type

The given equation \( y = \frac{1}{4}(x-2)^2 + 4 \) is a quadratic function, represented in vertex form \( y = a(x-h)^2 + k \), where \( (h, k) \) is the vertex.

Step 2: Determine the Vertex

In the function \( y = \frac{1}{4}(x-2)^2 + 4 \), the vertex \( (h, k) \) is \( (2, 4) \). This is because \( h = 2 \) and \( k = 4 \) in the vertex form equation.

Step 3: Analyze the Parabola

Since the coefficient \( a = \frac{1}{4} \), which is positive, the parabola opens upwards. The smaller magnitude of \( a \) (less than 1) means the parabola is wider than the standard \( y = x^2 \).

Step 4: Find Additional Points

Calculate a couple of points on either side of the vertex for accuracy in graphing. For example, substitute \( x = 0 \) to find \( y \), and substitute another value like \( x = 4 \).

Step 5: Calculate \( y \) for \( x = 0 \)

Substitute \( x = 0 \) into the equation: \[ y = \frac{1}{4}(0-2)^2 + 4 = \frac{1}{4} \times 4 + 4 = 1 + 4 = 5 \]. So, one point is \( (0, 5) \).

Step 6: Calculate \( y \) for \( x = 4 \)

Substitute \( x = 4 \) into the equation: \[ y = \frac{1}{4}(4-2)^2 + 4 = \frac{1}{4} \times 4 + 4 = 1 + 4 = 5 \]. So, another point is \( (4, 5) \).

Step 7: Plot the Points and Sketch the Parabola

Plot the vertex \( (2, 4) \) and the additional points \( (0, 5) \) and \( (4, 5) \). Then, draw a smooth curve passing through these points, ensuring the parabola opens upwards and symmetrically around the vertex.

Key Concepts

Parabola

A parabola is a U-shaped curve that can open either upwards or downwards. In the context of quadratic functions, it represents the graph of an equation in which the highest degree of the variable is 2.
The basic form of a quadratic equation is:

• \( y = ax^2 + bx + c \)

The shape and direction of the parabola depend on the coefficient \( a \).
If \( a \) is positive, the parabola opens upwards. If \( a \) is negative, it opens downwards. Additionally, the larger the absolute value of \( a \), the narrower the parabola. Conversely, the smaller the absolute value, the wider the parabola. In our function, \( a = \frac{1}{4} \), meaning the parabola opens upwards and is relatively wide because \( \frac{1}{4} \) is less than 1.
Understanding these basic characteristics will help you sketch and interpret parabolas easily.

Vertex Form

The vertex form of a quadratic equation is very handy for graphing and analyzing parabolas. It is expressed as:

• \(y = a(x-h)^2 + k\)

Here, \((h, k)\) represents the vertex of the parabola, a point where it reaches its maximum or minimum value depending on the direction it opens.
For our specific function \( y = \frac{1}{4}(x-2)^2 + 4 \), we can see the vertex at \((2, 4)\). This tells us not only the position of the center point but also that from here, the parabola will symmetrically extend on both sides.
Knowing the vertex off the bat simplifies determining the shape and position of the graph without further calculations.

Graphing Equations

Graphing a quadratic function involves a few steps to make sure the plot reflects the equation accurately. After identifying the vertex and the direction in which the parabola opens, you should calculate a few additional points.
These extra points help create a more accurate sketch. For example:

• Substituting specific \(x\) values into the equation can yield corresponding \(y\) points.
• In our example, for \(x = 0\) and \(x = 4\), the coordinates obtained were \((0, 5)\) and \((4, 5)\) respectively.

With these points and the vertex, which is \((2, 4)\), you can plot these on a graph.
A smooth curve connecting them depicts the parabola, completing the visual representation of the quadratic equation.
Remember, symmetry about the vertex is key in confirming your parabola's accuracy.