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Chapter 3: Problem 14

Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior. $$P(x)=(2 x-1)(x+1)(x+3)$$

Short Answer

Expert verified
The graph is a cubic curve with x-intercepts at -3, -1, 0.5 and y-intercept at -3.

Step by step solution

01

Identify the Polynomial's Degree

The given polynomial function is \( P(x)=(2x-1)(x+1)(x+3) \). To determine its degree, expand the expression. The highest degree of \( x \) indicates the degree of the polynomial. Multiplying all linear terms, \( (2x - 1)(x + 1)(x + 3) \), results in \( 2x^3 + 9x^2 + 4x - 3 \). Thus, the degree is 3, indicating a cubic polynomial.
02

Determine the End Behavior

For a cubic polynomial \( ax^3 + bx^2 + cx + d \) where \( a > 0 \), as \( x \to \infty, P(x) \to \infty \) and as \( x \to -\infty, P(x) \to -\infty \). Therefore, \( P(x) \) exhibits this behavior with \( a = 2 \). Thus, the graph will rise to the right and fall to the left.
03

Find the x-intercepts

The x-intercepts occur where \( P(x) = 0 \). By finding the roots of \( (2x-1)(x+1)(x+3) = 0 \), solve each factor for zero: \( 2x - 1 = 0 \) gives \( x = \frac{1}{2} \); \( x + 1 = 0 \) gives \( x = -1 \); \( x + 3 = 0 \) gives \( x = -3 \). So, the x-intercepts are at \( x = \frac{1}{2}, -1, -3 \).
04

Find the y-intercept

To find the y-intercept, evaluate \( P(x) \) at \( x = 0 \). \( P(0) = (2(0)-1)((0)+1)((0)+3) = -1 \cdot 1 \cdot 3 = -3 \). Thus, the y-intercept is \( -3 \).
05

Sketch the Graph

Using the information gathered, first plot the intercepts: x-intercepts at \( x = -3, -1, \frac{1}{2} \) and the y-intercept at \( y = -3 \). Considering the end behavior (fall to the left, rise to the right), sketch the curve passing through the intercepts, starting from below the x-axis near \( x = -\infty \), crossing the x-axis at \( x = -3, -1, \frac{1}{2} \), and finally rising above the x-axis as \( x \to \infty \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cubic Polynomials
Cubic polynomials are algebraic expressions where the highest exponent of the variable, usually denoted by \( x \), is three. In simple terms, it means that the equation involves \( x^3 \) as the term with the highest power. These equations can take a general form of \( ax^3 + bx^2 + cx + d \), where \( a \), \( b \), \( c \), and \( d \) are constants.
  • The term \( ax^3 \) influences the polynomial鈥檚 end behavior dramatically.
  • Cubic equations can have up to three real roots which translate into x-intercepts on a graph.
  • The shape of a cubic polynomial graph is typically an 'S' or 'N' like curve due to its degree.
Understanding the structure of a cubic polynomial, like in the exercise where \( P(x) = (2x - 1)(x + 1)(x + 3) \) was expanded to \( 2x^3 + 9x^2 + 4x - 3 \), helps in predicting its behavior and features like turning points and intercepts.
End Behavior
The end behavior of a polynomial describes how the graph behaves as \( x \) approaches infinity, whether positive or negative. For cubic polynomials, this behavior largely depends on the leading coefficient, which is the coefficient of the term with the highest degree, \( ax^3 \).
  • If \( a > 0 \), the right side of the graph will rise to infinity as \( x \rightarrow \infty \) and fall to negative infinity as \( x \rightarrow -\infty \).
  • If \( a < 0 \), the graph will fall to negative infinity as \( x \rightarrow \infty \) and rise to positive infinity as \( x \rightarrow -\infty \).
In our exercise, \( P(x) = 2x^3 + 9x^2 + 4x - 3 \) has a leading term of \( 2x^3 \) with \( a = 2 \), which means the graph rises on the right and falls on the left. This predictable behavior allows us to sketch the graph accurately.
X-Intercepts
X-intercepts occur where the graph of the polynomial crosses the x-axis. This happens when \( P(x) = 0 \). These intercepts are also referred to as roots or solutions of the polynomial equation. For the polynomial \( P(x) = (2x - 1)(x + 1)(x + 3) \), setting each factor equal to zero helps in finding the x-intercepts.
  • For \( 2x - 1 = 0 \), solve for \( x \) to get \( x = \frac{1}{2} \).
  • For \( x + 1 = 0 \), solve for \( x \) to get \( x = -1 \).
  • For \( x + 3 = 0 \), solve for \( x \) to get \( x = -3 \).
Therefore, the graph will cross the x-axis at \( x = -3, -1, \) and \( \frac{1}{2} \). Marking these points on the graph is an effective way to understand the polynomial鈥檚 real roots and guide the overall shape of the curve.
Y-Intercept
The y-intercept is the point where the polynomial graph crosses the y-axis. To find this, you simply evaluate the function at \( x = 0 \). This means setting \( x \) to zero in the equation and solving for \( y \).
For our polynomial function \( P(x) = (2x - 1)(x + 1)(x + 3) \), calculate:
  • \( P(0) = (2(0) - 1)((0) + 1)((0) + 3) \).
  • Simplifying the expression gives \( -1 \cdot 1 \cdot 3 = -3 \).
Therefore, the y-intercept is at \( y = -3 \). On the graph, this is the point where the curve touches the y-axis, and it helps in determining the initial path of the curve when sketching the polynomial.

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Most popular questions from this chapter

The real solutions of the given equation are rational. List all possible rational roots using the Rational Zeros Theorem, and then graph the polynomial in the given viewing rectangle to determine which values are actually solutions. (All solutions can be seen in the given viewing rectangle.) $$3 x^{3}+8 x^{2}+5 x+2=0 ; \quad[-3,3] \text { by }[-10,10]$$

Find the intercepts and asymptotes, and then sketch a graph of the rational function. Use a graphing device to confirm your answer. $$t(x)=\frac{x^{3}-x^{2}}{x^{3}-3 x-2}$$

How Many Real Zeros Can a Polynomial Have? Give examples of polynomials that have the following properties, or explain why it is impossible to find such a polynomial. (a) A polynomial of degree 3 that has no real zeros (b) A polynomial of degree 4 that has no real zeros (c) A polynomial of degree 3 that has three real zeros, only one of which is rational (d) A polynomial of degree 4 that has four real zeros, none of which is rational What must be true about the degree of a polynomial with integer coefficients if it has no real zeros?

Graph the rational function and find all vertical asymptotes, \(x\)- and \(y\)-intercepts, and local extrema, correct to the nearest decimal. Then use long division to find a polynomial that has the same end behavior as the rational function, and graph both functions in a sufficiently large viewing rectangle to verify that the end behaviors of the polynomial and the rational function are the same. $$r(x)=\frac{x^{4}-3 x^{3}+6}{x-3}$$

Find all rational zeros of the polynomial, and then find the irrational zeros, if any. Whenever appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartes' Rule of Signs, the quadratic formula, or other factoring techniques. $$P(x)=2 x^{4}+3 x^{3}-4 x^{2}-3 x+2$$
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