Problem 67 - College Algebra with Corequisite Support

Question

For the following exercises, write a formula for the function g that results when the graph of a given toolkit function is transformed as described. The graph of \(f(x)=x^{2}\) is vertically compressed by a factor of \(\frac{1}{2},\) then shifted to the right 5 units and up 1 unit.

Step-by-Step Solution

Answer

\( g(x) = \frac{1}{2}(x-5)^2 + 1 \)

Step 1: Identify the Toolkit Function

The given toolkit function is the quadratic function \( f(x) = x^2 \). This function forms a parabola with its vertex at the origin.

Step 2: Apply Vertical Compression

To vertically compress the graph of \( f(x) = x^2 \) by a factor of \( \frac{1}{2} \), we multiply the entire function by \( \frac{1}{2} \). This gives us \( g(x) = \frac{1}{2} x^2 \).

Step 3: Translate to the Right

To shift the graph 5 units to the right, we replace \( x \) with \( x-5 \) in the function. The new function becomes \( g(x) = \frac{1}{2}(x-5)^2 \).

Step 4: Translate Upward

To shift the graph up by 1 unit, we add 1 to the entire function. This modifies the function to \( g(x) = \frac{1}{2}(x-5)^2 + 1 \).

Step 5: Write the Final Function

Combining all transformations, the final function is \( g(x) = \frac{1}{2}(x-5)^2 + 1 \).

Key Concepts

Quadratic Functions

Quadratic functions are fundamental in mathematics. They are expressed as \( f(x) = ax^2 + bx + c \). This formula represents a parabola, a U-shaped curve. A special property of parabolas is that they have a vertex, which is their highest or lowest point, depending on whether they open upwards or downwards. For the simplest quadratic function, \( f(x) = x^2 \), the vertex is at the origin \((0,0)\). This function is symmetric about the y-axis. Knowing this basic shape helps understand how transformations affect it.

• Standard form: \( f(x) = ax^2 + bx + c \)
• Vertex form: \( f(x) = a(x-h)^2 + k \)

Quadratic functions show us how values of \( x \) are squared and what impact this has graphically.

Vertical Compression

Vertical compression affects the steepness of a quadratic function. It occurs when you multiply the function by a factor between 0 and 1. In the given example, we used a factor of \( \frac{1}{2} \). This makes the parabola wider and less steep compared to the original. Think of this transformation as squishing the graph vertically.

• If the factor is the same on both sides of the y-axis, the curve remains symmetric, just spread out.
• When \( a = 1 \), the parabola is normal; if less than 1, it's compressed.

By applying a vertical compression to \( f(x) = x^2 \), we got the modified function \( g(x) = \frac{1}{2}x^2 \). This transformation doesn't affect the position of the vertex but changes how sharp the curve is.

Horizontal Shift

A horizontal shift changes the position of the graph along the x-axis. To achieve this, you replace \( x \) with \( x - h \), where \( h \) is the number of units you want to move the graph.

• To shift right, decrease the value by \( h \), i.e., \( x - h \).
• To shift left, increase the value, i.e., \( x + h \).

For our example, the graph of \( \frac{1}{2}x^2 \) was shifted 5 units right, using \( (x-5) \). Thus, we expressed it as \( g(x) = \frac{1}{2}(x-5)^2 \). This means the entire graph moves right, changing the vertex to \((5,0)\).

Vertical Shift

Vertical shifts adjust the position of a graph along the y-axis. To apply a vertical shift, you add or subtract a value directly to or from the function.

• Shift up by adding a constant \( k \).
• Shift down by subtracting a constant \( k \).

In this exercise, after horizontal shift and vertical compression, the function was moved up 1 unit. The new expression became \( g(x) = \frac{1}{2}(x-5)^2 + 1 \). The result is that each point on the graph, including the vertex, elevates by 1 unit. Consequently, the vertex moves vertically to \((5,1)\). This transformation merely changes where the graph sits without altering its shape.