Problem 32 - Precalculus : Building Concepts and Connections

Question

For each polynomial function, find (a) the end behavior; (b) the \(y\) -intercept; (c) the \(x\) -intercept(s) of the graph of the function and the multiplicities of the real zeros; (d) the symmetries of the graph of the function, if any; and (e) the intervals on which the function is positive or negative. Use this information to sketch a graph of the function. Factor first if the expression is not in factored form. $$f(x)=(x-1)(x+2)^{2}(x+1)$$

Step-by-Step Solution

Answer

The end behavior is: \(x \to \pm \infty, f(x) \to +\infty\). The \(y\)-intercept is \(-4\). The \(x\)-intercepts are \(1, -2, -1\), with multiplicities \(1, 2, 1\), respectively. There are no symmetries. The interval where the function is positive is \((-2, -1)\) union \((1, \infty)\) and where it's negative is \((-\infty, -2)\) union \((-1, 1)\). Graph the function with these properties.

Step 1: Find the End Behavior

The end behavior of a function is determined by looking at the leading term when it is written in standard form. For the given function \(f(x)=(x-1)(x+2)^{2}(x+1)\), the leading term when multiplied out is \(x^{4}\), which is a positive coefficient and an even degree. Therefore, as \(x\) approaches positive or negative infinity, \(f(x)\) also approaches positive infinity.

Step 2: Find the \(y\)-Intercept

The \(y\)-intercept of the graph is the function's value at \(x=0\). Substitute \(0\) into \(f(x)\) to find the \(y\)-intercept. \(f(0) = (0-1)(0+2)^{2}(0+1) = -4\)

Step 3: Find the \(x\)-Intercepts

The \(x\)-intercepts of the graph are the real roots of the function. Set the function equal to \(0\) to find the solutions. \(0=(x-1)(x+2)^{2}(x+1)\) gives \(x = 1, -2, -1\). The multiplicity of the real zero \(x = -2\) is \(2\) because \((x+2)^{2}\) is a factor, and the multiplicities of the real zeros \(x = 1, -1\) is \(1\)

Step 4: Identify the Symmetries of the Graph

Check if the function is even, odd, or neither. \(f(-x)\) and \(f(x)\) are not identical and not negatives of each other, hence the function is neither even nor odd. No symmetry exists.

Step 5: Determine the Intervals of Positivity and Negativity

Plot the \(x\)-intercepts on a number line. Test a number from each interval in the function. \(f(x) > 0\) when \(x\) is in \((-2, -1)\) or \((1, \infty)\) and \(f(x) < 0\) when \(x\) is in \((-\infty, -2)\) or \((-1, 1)\)

Step 6: Sketch the Graph of the Function

Using all the information collected, one can now sketch a graph. The graph starts from the upper end because of the end behavior. It crosses or touches the \(x\)-axis at the \(x\)-intercepts. It is above or below the \(x\)-axis where the function is positive or negative respectively. There are no symmetries.

Key Concepts

End Behavior

The end behavior of a polynomial function refers to how the function behaves as the input, \( x \), approaches positive or negative infinity. To determine this, we look at the leading term of the polynomial when it is expressed in its expanded form.

For our given function \( f(x)=(x-1)(x+2)^2(x+1) \), the leading term is \( x^4 \). This term is derived by multiplying all the highest degree terms from each factor.

- **Degree**: Since the degree is four and an even number, this indicates that both ends of the graph will have similar behavior.- **Leading coefficient**: Here, the leading coefficient is positive (1), meaning that as \( x \to \infty \) or \( x \to -\infty \), the function \( f(x) \to \infty \).

Simply put, the graph of the function will rise to infinity on both sides.

Intercepts

Intercepts are essential in graphing polynomial functions as they provide key points where the graph crosses the axes. The two kinds of intercepts typically considered are the \( y \)-intercept and the \( x \)-intercepts.

To find the **\( y \)-intercept**, we substitute \( x = 0 \) into the function \( f(x) \). For \( f(x)=(x-1)(x+2)^2(x+1) \), this gives:

• \( f(0) = (0-1)(0+2)^2(0+1) = -4 \)

Thus, the graph crosses the \( y \)-axis at \((0, -4)\).

Let's move on to the **\( x \)-intercepts**. These occur where the graph crosses the \( x \)-axis, meaning \( f(x) = 0 \). For our function:

• \( x = 1 \)
• \( x = -2 \) (with multiplicity 2)
• \( x = -1 \)

The multiplicity of \( x = -2 \) means the graph simply "bounces off" the axis here, rather than crossing it.

Graph Symmetries

Symmetries in a graph can simplify our understanding and sketching of the polynomial function. There are two primary types of symmetry to check for: **even symmetry** and **odd symmetry**.

- **Even Symmetry**: A function is even if \( f(-x) = f(x) \) for all \( x \). This means the graph will be symmetrical about the \( y \)-axis.- **Odd Symmetry**: A function is odd if \( f(-x) = -f(x) \) for all \( x \), resulting in symmetry about the origin.

For our polynomial function \( f(x)=(x-1)(x+2)^2(x+1) \), both conditions for even and odd symmetries are not satisfied. Evaluating \( f(-x) \) does not produce \( f(x) \) nor \( -f(x) \). Hence, this function has no symmetry with respect to the \( y \)-axis or the origin.

Intervals of Positivity and Negativity

The intervals where the function is positive or negative are crucial to understanding its graph behavior. These intervals are defined based on where the graph lies relative to the \( x \)-axis. To find them, follow these steps:

Plot the \( x \)-intercepts on a number line (i.e., \( x = 1, -2, -1 \)). These points divide the number line into intervals:

• \((-\infty, -2)\)
• \((-2, -1)\)
• \((-1, 1)\)
• \((1, \infty)\)

By testing a point from each interval, you can ascertain where the function is positive or negative:

• \( f(x) < 0 \) in \((-\infty, -2)\) and \((-1, 1)\)
• \( f(x) > 0 \) in \((-2, -1)\) and \((1, \infty)\)

This behavior helps sketching the graph, as you can draw above the \( x \)-axis when the function is positive and below when it is negative.

Real Zeros Multiplicity

Real zeros, often known as roots, of a polynomial are the \( x \)-values where the function equals zero, while their multiplicities indicate how the graph interacts with the \( x \)-axis at those points.

Consider the function \( f(x) = (x-1)(x+2)^2(x+1) \):

• \( x = 1 \) has a multiplicity of 1, implying the graph crosses the \( x \)-axis directly.
• \( x = -2 \) has a multiplicity of 2, so the graph "touches" and bounces off the \( x \)-axis without crossing it.
• \( x = -1 \) also has a multiplicity of 1, where the graph crosses the \( x \)-axis.

The concept of multiplicity is vital as it affects both the shape of the graph and how it interacts with the \( x \)-axis at each root. Higher multiplicities mean a flattening and bouncing of the graph at those roots.