Graphing Polynomials

Graph polynomials ; a step by step tutorial with examples and detailed solutions. Factoring, zeros and their multiplicities, intercepts and other properties are used to graph polynomials.

Examples with Detailed Solutions

Example 1

Solution to Example 1

• a) Factor \( P \) as follows: \[ P(x) = -x^3 - x^2 + 2x \] \[ = -x(x^2 + x - 2) \] \[ = -x(x + 2)(x - 1) \]
• b) \( P \) has three zeros which are \( -2 \), \( 0 \), and \( 1 \), all with multiplicity one.
• c) The three zeros of \( P \) split the number line into four intervals: \[ (-\infty, -2), \quad (-2, 0), \quad (0, 1), \quad (1, +\infty) \] Select one value of \( x \) within each interval and evaluate \( P \) at that value to determine the sign of \( P \).
\[ P(-3) = 12 > 0, \quad P(-1) = -2 < 0, \quad P\left(\frac{1}{2}\right) = \frac{5}{8} > 0, \quad P(2) = -8 < 0 \] Using the signs of \( P \) in each interval, the sign chart is as follows:

Example 2

Solution to Example 2

• a) Factor \( P \) as follows \[ P(x) = x^4 - 2x^2 + 1 \] \[ = (x^2 - 1)^2 \] \[ = ((x - 1)(x + 1))^2 \] \[ = (x - 1)^2(x + 1)^2 \]
• b) Polynomial \( P \) has zeros at \( x = 1 \) and \( x = -1 \), both with multiplicity 2.
• c) Polynomial \( P(x) \) is a perfect square and therefore nonnegative for all real values of \( x \). \( P(x) \) is equal to zero at the two zeros \( -1 \) and \( 1 \), and positive everywhere else. The sign chart is as follows:

  

• d) The \( x \)-intercepts are at \( (-1, 0) \) and \( (1, 0) \), and the \( y \)-intercept is at \( (0, 1) \). The graph of \( P \) touches the \( x \)-axis at \( x = -1 \) and \( x = 1 \), and opens upward since \( P(x) \) is positive, cutting the \( y \)-axis at \( (0, 1) \). See graph below:

  

• e) Using the graph of \( P \) above, the range of \( P \) is given by the interval \[ [0, +\infty) \]

Example 3

c) Factor P and determine the multiplicity of each zero of \( P \).

d) Determine the sign chart of \( P \).

e) Graph polynomial P and label the x and y intercepts on the graph obtained.

Solution to Example 3

• a) Calculate \( P(-3) \) \[ P(-3) = (-3)^4 + 5(-3)^3 + 5(-3)^2 - 5(-3) - 6 = 0 \] Hence \(-3\) is a zero of \(P\), and \(x + 3\) is a factor of \(P(x)\).
• b) Since \(x + 3\) is a factor of \(P(x)\), the division of \(P(x)\) by \(x + 3\) must give a remainder equal to zero. Hence \[ \frac{P(x)}{x + 3} = x^3 + 2x^2 - x - 2 \] We now divide \(x^3 + 2x^2 - x - 2\) by \(x - 1\): \[ \frac{x^3 + 2x^2 - x - 2}{x - 1} = x^2 + 3x + 2 \] The remainder is 0, which proves that \(x - 1\) is a factor of \(x^3 + 2x^2 - x - 2\), and therefore also of \(P(x)\).

• c) Using the above, \(P(x)\) may be written as: \[ P(x) = (x + 3)(x - 1)(x^2 + 3x + 2) \] We now factor the quadratic term \(x^2 + 3x + 2\) included in \(P(x)\). Hence: \[ P(x) = (x + 3)(x - 1)(x + 1)(x + 2) \] \(P\) has zeros at \(x = -3\), \(-2\), \(-1\), and \(1\), and all are of multiplicity one.
• d) The sign chart is shown below:

  

• e) Using the information on the zeros and the sign chart, the graph of \(P\) is as shown below with \(x\)- and \(y\)-intercepts labeled.

  

Example 4

Solution to Example 4

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