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For Problems \(35-42\), (a) find the \(y\) intercepts, (b) find the \(x\) intercepts, and (c) find the intervals of \(x\) where \(f(x)>0\) and those where \(f(x)<0\). Do not sketch the graphs. $$ f(x)=(x-4)^{2}(x+3)^{3} $$

Short Answer

Expert verified
y-intercept: (0, 432); x-intercepts: (4, 0), (-3, 0); f(x) > 0 for x ∈ (-∞, -3) and (4, ∞); f(x) < 0 for x ∈ (-3, 4).

Step by step solution

01

Find the y-intercept

To find the y-intercept of the function, evaluate the function at \(x = 0\). This means substituting \(x = 0\) into \(f(x) = (x-4)^{2}(x+3)^{3}\): \[ f(0) = (0 - 4)^{2}(0 + 3)^{3} = (16)(27) = 432. \] Hence, the y-intercept is \((0, 432)\).
02

Find the x-intercepts

To find the x-intercepts, set the function equal to zero and solve for \(x\): \((x-4)^{2}(x+3)^{3} = 0\). This occurs when either \((x-4)^{2} = 0\) or \((x+3)^{3} = 0\). Solving these gives: \[ x-4 = 0 \Rightarrow x = 4 \] \[ x+3 = 0 \Rightarrow x = -3 \] Thus, the x-intercepts are \((4, 0)\) and \((-3, 0)\).
03

Determine intervals where \(f(x) > 0\) and \(f(x) < 0\)

To determine where the function is positive or negative, analyze the sign of \(f(x)\) between and beyond the x-intercepts, at \(x = -3\) and \(x = 4\). **Intervals:** \((-\infty, -3)\), \((-3, 4)\), \((4, \infty)\). - For \(x < -3\), both \(x-4\) and \(x+3\) are negative, so \(f(x) > 0\). - For \(-3 < x < 4\), \(x-4\) is negative and \(x+3\) is positive, so \(f(x) < 0\). - For \(x > 4\), both \(x-4\) and \(x+3\) are positive, so \(f(x) > 0\). Thus, \(f(x) > 0\) on \((-\infty, -3)\) and \((4, \infty)\), and \(f(x) < 0\) on \((-3, 4)\).

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Key Concepts

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

Understanding Y-intercepts
The y-intercept of a polynomial function is where the graph crosses the y-axis. To find the y-intercept, we substitute zero for every x in the function, since the y-axis is at x = 0. For our function, \( f(x) = (x-4)^2(x+3)^3 \), we calculate \( f(0) \). This requires evaluating the expression by substituting \( x = 0 \) into the function, resulting in:\[f(0) = (0 - 4)^2 (0 + 3)^3 = 16 \times 27 = 432.\]
The calculation shows the y-intercept to be (0, 432). This point tells us where the curve touches or crosses the y-axis on a graph.
Finding X-intercepts
X-intercepts occur where the graph of the function crosses the x-axis. This happens when the output of the function, \( f(x) \), equals zero. For our polynomial, set the expression to zero:\[(x-4)^2(x+3)^3 = 0\]
Solve for x by setting each factor equal to zero individually:
  • \( (x-4)^2 = 0 \) gives \( x = 4 \)
  • \( (x+3)^3 = 0 \) gives \( x = -3 \)
Thus, the x-intercepts are \((4, 0)\) and \((-3, 0)\). Each intersept indicates the function meets the x-axis at these coordinates, which are pivotal when analyzing the graph's shape and behavior.
Analyzing Function Intervals
Analyzing intervals involves understanding different parts of the number line and how the function behaves in each segment. Consider the x-intercepts at \( x = -3 \) and \( x = 4 \), which break the x-axis into distinct intervals:
  • \(( -\infty, -3 )\)
  • \( (-3, 4) \)
  • \( (4, \infty) \)
In essence:
- Between \( ( -\infty, -3 ) \), substitute any point in this range into \( (x-4) \) and \( (x+3) \). Both factors are negative here, meaning their product is positive.- Within \( (-3, 4) \), the terms \( (x-4) \) becomes negative while \( (x+3) \) remains positive, yielding a negative outcome.- Beyond \( (4, \infty) \), both parts \( (x-4) \) and \( (x+3) \) are positive, resulting in positive values.
These behaviors guide us in identifying important properties of the function, necessary for sketching graphs or understanding real-world applications.
Positivity and Negativity of Functions
To identify positivity and negativity of the function \( f(x) = (x-4)^2(x+3)^3 \), observe where the function results in positive or negative values. Such examination depends on the sign of the function over determined intervals:
**Positive intervals** occur where the product of the factors yields a positive value. For example:
  • **\(( -\infty, -3 )\):** Both factors are negative, resulting in a positive product.
  • **\(( 4, \infty )\):** Both factors are positive, resulting in a positive product.
**Negative intervals** are characterized by a negative product of the factors. Between:
  • **\(( -3, 4 )\):** The mixture of one negative and one positive factor outcomes in a negative product.
These intervals reflect where the graph is positioned relative to the x-axis (above or below), offering insight into the reality of solutions or roots that might exist for applied scenarios.

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Most popular questions from this chapter

Suppose that \(x\) ounces of pure acid have been added to 14 ounces of a \(15 \%\) acid solution. (a) Set up the rational expression that represents the concentration of pure acid in the final solution. (b) Graph the rational function that displays the concentration. (c) How many ounces of pure acid need to be added to the 14 ounces of a \(15 \%\) solution to raise it to a \(40.5 \%\) solution? Check your answer. (d) How many ounces of pure acid need to be added to the 14 ounces of a \(15 \%\) solution to raise it to a \(50 \%\) solution? Check your answer. (e) What concentration of acid do we obtain if we add 12 ounces of pure acid to the 14 ounces of a \(15 \%\) solution? Check your answer.

Find the coordinates of the turning points of each of the following graphs. Express \(x\) and \(y\) values to the nearest integer. (a) \(f(x)=2 x^{3}-3 x^{2}-12 x+40\) (b) \(f(x)=2 x^{3}-33 x^{2}+60 x+1050\) (c) \(f(x)=-2 x^{3}-9 x^{2}+24 x+100\) (d) \(f(x)=x^{4}-4 x^{3}-2 x^{2}+12 x+3\) (e) \(f(x)=x^{3}-30 x^{2}+288 x-900\) (f) \(f(x)=x^{5}-2 x^{4}-3 x^{3}-2 x^{2}+x-1\)

For Problems \(1-22\), graph each of the polynomial functions. $$ f(x)=(x-1)(x+1)(x-3) $$

Graph each of the following rational functions: $$ f(x)=\frac{3}{(x+2)(x-4)} $$

A polynomial function with real coefficients is continuous everywhere; that is, its graph has no holes or breaks. This is the basis for the following property: If \(f(x)\) is a polynomial with real coefficients, and if \(f(a)\) and \(f(b)\) are of opposite sign, then there is at least one real zero between \(a\) and \(b\). This property, along with our knowledge of polynomial functions, provides the basis for locating and approximating irrational solutions of a polynomial equation. Consider the equation \(x^{3}+2 x-4=0\). Applying Descartes' rule of signs, we can determine that this equation has one positive real solution and two nonreal complex solutions. (You may want to confirm this!) The rational root theorem indicates that the only possible rational solutions are 1,2 , and 4 . Using a little more compact format for synthetic division, we obtain the following results when testing for 1 and 2 as possible solutions: $$ \begin{array}{r|rrrr} & 1 & 0 & 2 & -4 \\ 1 & 1 & 1 & 3 & -1 \\ 2 & 1 & 2 & 6 & 8 \end{array} $$ Because \(f(1)=-1\) and \(f(2)=8\), there must be an irrational solution between 1 and 2 . Furthermore, \(-1\) is closer to 0 than is 8 , so our guess is that the solution is closer to 1 than to 2 . Let's start looking at \(1.0,1.1,1.2\), and so on, until we can place the solution between two numbers. Because \(f(1.1)=-0.469\) and \(f(1.2)=0.128\), the irrational solution must be between \(1.1\) and 1.2. Furthermore, because \(0.128\) is closer to 0 than is \(-0.469\), our guess is that the solution is closer to \(1.2\) than to \(1.1\). Let's start looking at \(1.15,1.16\), and so on. $$ \begin{array}{l|rrrrr} & 1 & 0 & 2 & -4 \\ \ 1.15 & 1 & 1.15 & 3.3225 & -0.179 \\ 1.16 & 1 & 1.16 & 3.3456 & -0.119 \\ 1.17 & 1 & 1.17 & 3.3689 & -0.058 \\ 1.18 & 1 & 1.18 & 3.3924 & 0.003 \end{array} $$ Because \(f(1.17)=-0.058\) and \(f(1.18)=0.003\), the irrational solution must be between \(1.17\) and \(1.18\). Therefore we can use \(1.2\) as a rational approximation to the nearest tenth. For each of the following equations, (a) verify that the equation has exactly one irrational solution, and (b) find an approximation, to the nearest tenth, of that solution. (a) \(x^{3}+x-6=0 \) (b) \(x^{3}-6 x-6=0 \) (c) \(x^{3}-27 x-60=0 \) (d) \(x^{3}-x^{2}-x-1=0 \) (e) \(x^{3}-2 x-10=0\) (f) \(x^{3}-5 x^{2}-1=0 \)

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