/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 (a) Use a graphing utility to dr... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Use a graphing utility to draw a graph of each function. (b) For each \(x\) -intercept, zoom in until you can estimate it accurately to the nearest one-tenth. (c) Use algebra to determine each \(x\) -intercept. If an intercept involves a radical, give that answer as well as a calculator approximation rounded to three decimal places. Check to see that your results are consistent with the graphical estimates obtained in part (b). $$W(u)=2 u^{4}-17 u^{2}+35$$

Short Answer

Expert verified
The x-intercepts are approximately \( -2.236, -1.871, 1.871, \) and \( 2.236 \).

Step by step solution

01

Understand the Function

We are given the function \( W(u) = 2u^4 - 17u^2 + 35 \). This is a polynomial of degree 4.
02

Graph the Function

Use a graphing calculator or online graphing tool to plot the function \( W(u) = 2u^4 - 17u^2 + 35 \). Observe where the function crosses the x-axis. These points are the x-intercepts.
03

Zoom and Estimate the x-intercepts

Zoom into each point where the graph crosses the x-axis to estimate the x-intercepts to the nearest one-tenth. Note these approximate values for comparison in later steps.
04

Find x-intercepts Algebraically

Set \( W(u) = 0 \): \[ 2u^4 - 17u^2 + 35 = 0 \]Let \( x = u^2 \), then the equation becomes \( 2x^2 - 17x + 35 = 0 \). This quadratic can be factored or solved using the quadratic formula.
05

Solve the Quadratic Equation

Factor \( 2x^2 - 17x + 35 \): \((2x - 7)(x - 5) = 0 \).Thus, solutions for \( x \) are \( x = \frac{7}{2} \) and \( x = 5 \). Convert back to \( u \): \[ u^2 = \frac{7}{2} \Rightarrow u = \pm\sqrt{\frac{7}{2}} \]\[ u^2 = 5 \Rightarrow u = \pm\sqrt{5} \].
06

Provide Approximations

Calculate the decimal approximations:\( \sqrt{\frac{7}{2}} \approx 1.871 \) and \( -\sqrt{\frac{7}{2}} \approx -1.871 \).\( \sqrt{5} \approx 2.236 \) and \( -\sqrt{5} \approx -2.236 \).
07

Compare Algebraic Results with Graph

Check that these approximations match the visual x-intercepts observed in Step 3 to verify the solutions are consistent with the graph.

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Key Concepts

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

x-intercepts
In math, the x-intercepts of a function are the points where its graph crosses the x-axis. At these points, the value of the function is zero. They are critical for understanding how a function behaves as these points are essentially the solutions to the equation.

To find the x-intercepts of a polynomial function like \( W(u) = 2u^4 - 17u^2 + 35 \), we set the function equal to zero and solve the resulting equation:
  • Step 1: Set \( W(u) = 0 \).
  • Step 2: Solve the equation for \( u \) to find the x-intercepts.
These steps help us to identify the u-values where the graph intersects the x-axis. In this example, the intercepts are found by converting the quartic equation into a quadratic one, simplifying the process. Ensure mathematical accuracy by checking these with graphing approximations.
quartic polynomial
A quartic polynomial is a polynomial of degree four, which means the highest power of the variable is four. These polynomials can be complex in nature, but understanding their behavior is facilitated by recognizing key patterns. Given the function \( W(u) = 2u^4 - 17u^2 + 35 \), we note that it is a quartic polynomial because of the \( u^4 \) term.

Quartic polynomials like this one can have zero to four x-intercepts. They might exhibit a variety of curve shapes, which can include turning points or regions where the graph neither crosses nor touches the x-axis. Simplifying a quartic equation often involves factoring it into a lower degree polynomial, in this case, a quadratic. This step involves recognizing or manipulating the polynomial structure to solve for roots in an easier form.
graphing utilities
Graphing utilities, such as a graphing calculator or online plotting tools, are essential tools for visualizing and analyzing polynomial functions. They provide a visual representation of the function's behavior and allow for approximation and verification of solutions.

Using graphing utilities with our given function \( W(u) = 2u^4 - 17u^2 + 35 \):
  • You can quickly identify where the graph crosses the x-axis by plotting the function.
  • Once the graph is plotted, you can use zoom features to closely examine the x-intercepts, approximating their values.
  • Interactive tools help refine these approximations, making it easier to check against calculations derived algebraically.
Modern graphing utilities are powerful for supporting and validating algebraic methods, especially in complex cases like quartic polynomials.
quadratic solutions
In cases where quartic polynomials can be simplified to quadratics, understanding quadratic solutions becomes crucial. The simplified quadratic equation from the problem is \( 2x^2 - 17x + 35 = 0 \). Quadratic equations can be solved using several techniques:
  • Factoring: This involves expressing the quadratic as a product of two binomials, finding the roots directly. In this case, \((2x - 7)(x - 5) = 0\).
  • Quadratic Formula: For quadratics that do not factor neatly, the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is used to find solutions.
From the quadratic roots, convert back to the original variable. Here, express solutions as \( u = \pm\sqrt{x} \). Calculating these roots gives both exact radical forms and approximate decimals, useful for comparison with graphing results.

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Most popular questions from this chapter

How far from the origin is the vertex of the parabola \(y=x^{2}-6 x+13 ?\)

(a) An open-top box is to be constructed from a 6 -by- 8 -in. rectangular sheet of tin by cutting out equal squares at each corner and then folding up the resulting flaps. Let \(x\) denote the length of the side of each cutout square. Show that the volume \(V(x)\) is $$V(x)=x(6-2 x)(8-2 x)$$ (b) What is the domain of the volume function in part (a)? [The answer is not \((-\infty, \infty) .]\) (c) Use a graphing utility to graph the volume function, taking into account your answer in part (b). (d) By zooming in on the turning point, estimate to the nearest one-hundredth the maximum volume.

Let \(f(x)=\left(x^{5}+1\right) / x^{2}\) (a) Graph the function \(f\) using a viewing rectangle that extends from -4 to 4 in the \(x\) -direction and from -8 to 8 in the \(y\) -direction. (b) Add the graph of the curve \(y=x^{3}\) to your picture in part (a). Note that as \(|x|\) increases (that is, as \(x\) moves away from the origin), the graph of \(f\) looks more and more like the curve \(y=x^{3} .\) For additional perspective, first change the viewing rectangle so that \(y\) extends from -20 to \(20 .\) (Retain the \(x\) -settings for the moment.) Describe what you see. Next, adjust the viewing rectangle so that \(x\) extends from -10 to 10 and \(y\) extends from -100 to \(100 .\) Summarize your observations. (c) In the text we said that a line is an asymptote for a curve if the distance between the line and the curve approaches zero as we move further and further out along the curve. The work in part (b) illustrates that a curve can behave like an asymptote for another curve. In particular, part (b) illustrates that the distance between the curve \(y=x^{3}\) and the graph of the given function \(f\) approaches zero as we move further and further out along the graph of \(f .\) That is, the curve \(y=x^{3}\) is an "asymptote" for the graph of the given function \(f\). Complete the following two tables for a numerical perspective on this. In the tables, \(d\) denotes the vertical distance between the curve \(y=x^{3}\) and the graph of \(f:\) $$ d=\left|\frac{x^{5}+1}{x^{2}}-x^{3}\right| $$ $$\begin{array}{llllll} \hline x & 5 & 10 & 50 & 100 & 500 \\ \hline d & & & & \\ \hline & & & & \\ \hline x & -5 & -10 & -50 & -100 & -500 \\ \hline d & & & & \\ \hline \end{array}$$ (d) Parts (b) and (c) have provided both a graphical and a numerical perspective. For an algebraic perspective that ties together the previous results, verify the following identity, and then use it to explain why the results in parts (b) and (c) were inevitable: $$ \frac{x^{5}+1}{x^{2}}=x^{3}+\frac{1}{x^{2}} $$

Determine the inputs that yield the minimum values for each function. Compute the minimum value in each case. (a) \(f(x)=\sqrt{x^{2}-6 x+73}\) (b) \(g(x)=\sqrt[3]{x^{2}-6 x+73}\) (c) \(h(x)=x^{4}-6 x^{2}+73\)

Find the distance between the vertices of the parabolas \(y=-\frac{1}{2} x^{2}+4 x\) and \(y=2 x^{2}-8 x-1\).

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