Problem 13 - College Algebra with Modeling and Visualization

Question

Find an equation that shifts the graph of \(f\) by the desired amounts. Do not simplify. Graph \(f\) and the shifted graph in the same \(xy\)-plane. \(f(x)=\frac{1}{2} x^{2}+2 x-1 ;\) left 3 units, downward 2 units

Step-by-Step Solution

Answer

The shifted equation is \( h(x) = \frac{1}{2}(x+3)^2 + 2(x+3) - 3 \).

Step 1: Analyze the original function

The original function is given as \( f(x) = \frac{1}{2}x^2 + 2x - 1 \). It is a quadratic function in the standard form \( ax^2 + bx + c \), where \( a = \frac{1}{2}, b = 2, \) and \( c = -1 \). The graph of this function is a parabola opening upwards.

Step 2: Determine horizontal shift

To shift the graph of the function to the left by 3 units, you replace \( x \) with \( x + 3 \) in the function. This results in the function \( g(x) = \frac{1}{2}(x+3)^2 + 2(x+3) - 1 \). This represents the horizontal shift.

Step 3: Determine vertical shift

For a downward shift of 2 units, you subtract 2 from the entire function. This results in the new function \( h(x) = \frac{1}{2}(x+3)^2 + 2(x+3) - 1 - 2 \). This is the function after both the leftward and downward shifts are applied.

Step 4: Construct the final shifted equation

Combine the transformations. The final equation after shifting the original \( f(x) \) left 3 units and down 2 units is \( h(x) = \frac{1}{2}(x+3)^2 + 2(x+3) - 3 \).

Key Concepts

Quadratic Functions

Quadratic functions are a crucial part of algebra and are represented by equations of the form \( ax^2 + bx + c \). These equations form a parabola when graphed on the Cartesian coordinate system. The general parabolic shape can either open upwards or downwards depending on the sign of the coefficient \( a \).

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), the parabola opens downwards.

In the given example, the original function is \( f(x) = \frac{1}{2}x^2 + 2x - 1 \), where \( a = \frac{1}{2} \), \( b = 2 \), and \( c = -1 \). Here, \( a \) is positive, indicating the parabola will open upwards. The importance of each coefficient is as follows:

• \( a \) determines the "width" of the parabola and the direction of opening.
• \( b \) influences the direction and location of the vertex horizontally.
• \( c \) shifts the parabola vertically on the graph.

Horizontal Shift

A horizontal shift in a function involves changing the input variable, \( x \), to produce a left or right movement of the graph. This is achieved by adjusting the \( x \) term inside the function. To move a graph left, you replace \( x \) with \( x + k \), and to shift it right, use \( x - k \), where \( k \) is the number of units to shift. In our problem, to shift the parabola left by 3 units, \( x \) is replaced by \( x + 3 \). So, for the function \( f(x) = \frac{1}{2}x^2 + 2x - 1 \), the new function becomes \( g(x) = \frac{1}{2}(x+3)^2 + 2(x+3) - 1 \). This transformation ensures that each point on the graph is moved three units to the left on the \( x \)-axis.Key points:

• The term inside the parenthesis \( x + 3 \) shows a leftward shift.
• No alterations are made to the surrounding coefficients or outside terms for horizontal shift.

Vertical Shift

In contrast to horizontal shifts, vertical shifts involve adding or subtracting a constant from the entire function. Such transformations move the graph up or down on the \( y \)-axis. To shift a graph down, subtract the desired number of units from the function. Conversely, to move it up, add units. Here, the original function was shifted down by 2 units. The transformation for our problem involves subtracting 2 from the function \( g(x) \). Therefore, the new function becomes \( h(x) = \frac{1}{2}(x+3)^2 + 2(x+3) - 3 \). This adjusted function accounts for the initial leftward shift and introduces the vertical shift.Helpful insights:

• The entire function \( g(x) \) is simply decreased by 2 to achieve the downward shift.
• Vertical shifts do not affect the shape or direction of the parabola, only its vertical position on the graph.