Problem 2 - Algebra 2 and Trigonometry

Question

Jordan said that if the roots of a polynomial function \(\mathrm{f}(x)\) are \(r_{1}, r_{2},\) and \(r_{3},\) then the roots of \(\mathrm{g}(x)=\mathrm{f}(x-a)\) are \(r_{1}+a, r_{2}+a,\) and \(r_{3}+a .\) Do you agree with Jordan? Explain why or why not.

Step-by-Step Solution

Answer

Yes, Jordan is correct. The roots of \( g(x)=f(x-a) \) are \( r_1+a, r_2+a, \) and \( r_3+a \).

Step 1: Understanding the Roots of f(x)

The given polynomial function \( f(x) \) has roots \( r_1, r_2, \) and \( r_3 \). This means \( f(r_1) = 0 \), \( f(r_2) = 0 \), and \( f(r_3) = 0 \).

Step 2: Substituting to Get g(x)

The function \( g(x) = f(x-a) \) involves a horizontal shift of the polynomial. To find its roots, we need \( f(x-a) = 0 \).

Step 3: Setting g(x) to Zero and Solving

For \( g(x) = 0 \), we need to solve \( f(x-a) = 0 \). Let \( u = x-a \), then \( f(u) = 0 \) which implies \( u = r_1, r_2, r_3 \).

Step 4: Rearranging to Solve for x

Since \( u = x-a \), we rearrange to find \( x = u + a \). This means \( x = r_1 + a, r_2 + a, r_3 + a \). Jordan's claim is correct, as applying this shift shows the new roots are indeed \( r_1 + a, r_2 + a, \) and \( r_3 + a \).

Key Concepts

Horizontal Shift in Functions

Let's explore what happens when we apply a horizontal shift to a function. Imagine you have a function \( f(x) \) with specific roots. A horizontal shift occurs when we replace \( x \) with \( x-a \). This shifts the graph horizontally to the right by \( a \) units if \( a \) is positive or to the left if \( a \) is negative.

In the context of polynomial roots, if \( f(x) \) has roots \( r_1, r_2, \) and \( r_3 \), the new function \( g(x) = f(x-a) \) will have roots that are shifted by the same amount, i.e., \( r_1+a, r_2+a, \) and \( r_3+a \). This is because solving \( g(x) = 0 \) leads to solving \( f(x-a) = 0 \). To find the values of \( x \) that satisfy this equation, we substitute \( x = u+a \) (where \( u = x-a \)) into the original root equations, leading to the conclusion that \( x \) equals the original roots plus \( a \).

• This shift is straightforward once you understand how the variable substitution affects the equation.
• It essentially moves the entire graph without altering its shape.

Transformation of Polynomial Functions

When we talk about transforming polynomial functions, it typically involves several types of changes, including shifts, stretches, and reflections.

The horizontal shift, as discussed, moves the graph without changing its overall form. This type of transformation is specific and only alters the position of the graph on the \( x \)-axis. However, transformations can also include vertical shifts and scaling (stretching or compressing the graph). For polynomial functions, transforming them might involve altering their coefficients or adding constants that affect their intercepts. This not only might shift the graph up or down, but it can also make the graph taller or shorter depending on the type of transformation applied.

• Vertical Shifts: Adding a constant \( b \) to \( f(x) \) moves the graph up or down.
• Scaling: Multiplying \( f(x) \) by a constant changes the steepness or flatness of the graph.

Whether through shifting or scaling, every transformation helps us manipulate polynomial functions to effectively fit or model different scenarios.

Solving Polynomial Equations

To solve polynomial equations, we typically aim to find values of \( x \) where \( f(x) = 0 \). These solutions are known as the roots of the polynomial. Knowing these roots is crucial for understanding the underlying behavior of the polynomial, as they indicate where the graph crosses the \( x \)-axis.

In our example, once we perform a horizontal shift to obtain \( g(x) = f(x-a) \), we also alter the polynomial equation that defines the graph's intersections. This new function requires a fresh approach to find its roots, yet the same principles apply:

• Find the roots of \( f(x-a) \) by considering equivalence \( f(u) = 0 \), where \( u = x-a \).
• Solve for \( x \) in terms of known roots \( u \).
• Rearrange back to express in terms of \( x \) using the found expressions.

Wisely working through these steps ensures that we maintain clear solutions, which reinforces Jordan’s statement. Each polynomial equation might appear complex, but focusing on its roots provides key insights into its complete structure.