Problem 118 - Algebra and Trigonometry

Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Write the factored form of the polynomial function of smallest degree that touches the \(x\) -axis at \(x=4,\) crosses the \(x\) -axis at \(x=-2\) and \(x=1,\) and has a \(y\) -intercept of 4.

Step-by-Step Solution

Answer

The factored form of the polynomial is \( f(x) = -\frac{1}{8}(x-4)^2(x+2)(x-1) \).

Step 1: Understanding the Problem

The polynomial function has roots where it crosses or touches the x-axis. According to the problem, the polynomial touches the x-axis at \(x=4\), crosses the x-axis at \(x=-2\) and \(x=1\), and has a y-intercept of 4.

Step 2: Identify the Roots of the Polynomial

Since the polynomial touches the x-axis at \(x=4\), this implies there is a double root at 4. Roots at \(x=-2\) and \(x=1\) imply simple roots. Therefore, the polynomial has roots at \(x=4\), \(x=4\), \(x=-2\), and \(x=1\).

Step 3: Write the Polynomial in Factored Form

Based on the identified roots, the polynomial can be written in factored form as: \[ f(x) = a(x-4)^2(x+2)(x-1) \] where \(a\) is a leading coefficient.

Step 4: Determine the Leading Coefficient Using the y-intercept

To find the leading coefficient \(a\), use the y-intercept given in the problem. When \(x=0\), \(f(x)=4\). Therefore, \[ f(0) = a(0-4)^2(0+2)(0-1) = 4 \] Solving for \(a\), \[ 4 = a(16)(2)(-1) \] \[ 4 = -32a \] \[ a = -\frac{1}{8} \]

Step 5: Write the Final Polynomial Function

Substitute \(a = -\frac{1}{8}\) back into the factored form of the polynomial: \[ f(x) = -\frac{1}{8}(x-4)^2(x+2)(x-1) \]

Key Concepts

Polynomial Roots

The roots of a polynomial are the values of x where the polynomial equals zero. These are also the points where the graph of the polynomial crosses or touches the x-axis. For example, in the given problem, the polynomial has roots at

• x = -2
• x = 1
• x = 4 (with a double root)

This means that the polynomial touches the x-axis at these points.

Double Root

A double root occurs when a polynomial touches the x-axis at a single point but doesn’t cross it. In mathematical terms, if a polynomial has a double root at x = 4, it means that (x-4)^2 is a factor of the polynomial. This gives the appearance of the graph 'just touching' the x-axis and turning around at that point.
This is different from simple roots, where the polynomial crosses the x-axis.

Factored Form of Polynomial

The factored form of a polynomial makes it easy to see its roots. It is written as a product of its factors, which include the double and simple roots. From the problem, the polynomial has roots at x = -2, x = 1, and x = 4. Therefore, the polynomial can be factored as: f(x) = a(x-4)^2(x+2)(x-1).
The 'a' represents a leading coefficient that adjusts the width and orientation of the graph.

Leading Coefficient

The leading coefficient in a polynomial is a crucial factor that determines the vertical stretch or compression of the graph. To find the leading coefficient 'a', you can use another point on the graph, such as the y-intercept. In our problem, the y-intercept is given as 4.\(f(0) = 4\).
By substituting x = 0 and solving for 'a', we find that: a = -\frac{1}{8}. This gives us the final polynomial function: f(x) = -\frac{1}{8}(x-4)^2(x+2)(x-1).