/*! 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 80 An equation is given. (a) Use a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An equation is given. (a) Use a graphing calculator to graph the equation in the given viewing rectangle. (b) Find the x- and y-intercepts from the graph. (c) Verify your answers to part (b) algebraically (from the equation). $$y=x^{4}-2 x^{3} ; \quad[-2,3] \text { by }[-3,3]$$

Short Answer

Expert verified
The x-intercepts are (0, 0) and (2, 0); the y-intercept is (0, 0).

Step by step solution

01

Understand the Graphing Task

We are given the polynomial equation \(y = x^4 - 2x^3\) and need to graph it using a graphing tool. We'll use the viewing rectangle \([-2, 3] \text{ by } [-3, 3]\). This means our x-values range from -2 to 3 and y-values from -3 to 3. Ensure the graphing calculator viewing window is set accordingly before plotting the equation.
02

Graph the Equation

Using a graphing calculator, enter the equation \(y = x^4 - 2x^3\) and plot graphically in the specified viewing window. Check that the graph appears fully within the rectangle. Observe the graph to gather information about the intercepts.
03

Identify the Intercepts Graphically

From the graph, locate where the curve crosses the x-axis and y-axis. The x-intercepts are where \(y=0\), and the y-intercept is where the graph crosses the y-axis. Note these intercept points: x-intercepts occur around the points where the curve cuts the x-axis, and the y-intercept occurs where the curve crosses the y-axis.
04

Verify Intercepts Algebraically

To find the intercepts algebraically, set the appropriate terms to zero. For the y-intercept, set \(x = 0\) in the equation: \(y = 0^4 - 2 \times 0^3 = 0\). Hence, the y-intercept is \((0, 0)\). For the x-intercepts, set the equation to zero and solve: \(x^4 - 2x^3 = 0\). Factor to get: \(x^3(x - 2) = 0\). This gives \(x = 0\) or \(x = 2\). Thus, the x-intercepts are \((0, 0)\) and \((2, 0)\).
05

Summarize Intercept Findings

Confirm that both graphical and algebraic approaches yield the same intercepts. The x-intercepts found are \((0, 0)\) and \((2, 0)\), and the y-intercept is \((0, 0)\). This aligns with our observations from the graph.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Graphing with Calculator
Graphing polynomial equations like \(y = x^4 - 2x^3\) can be simplified using a graphing calculator. These tools allow us to visualize complex equations quickly. To correctly graph an equation, first set the calculator's viewing window based on the specified ranges. Here, our x-values range from -2 to 3, and y-values from -3 to 3. Adjust your calculator settings to match this window, ensuring the graph will render correctly within the specified rectangle. Once set, input your equation into the calculator, and then graph it. This visual representation helps in analyzing the behavior of the equation, particularly where it meets the axes, indicating intercepts.

Seeing a graph can often clarify the equation's characteristics more than calculations alone, especially when visualizing polynomial curves.
X-Intercept
The x-intercept(s) of a graph are vital points where the graph crosses the x-axis, which means where \(y = 0\). These points provide insight into the equation's roots or real solutions. Graphically, you can identify x-intercepts by looking at where the curve cuts the x-axis on the graphing calculator's display.

For the equation \(y = x^4 - 2x^3\), the graph shows x-intercepts at \((0, 0)\) and \((2, 0)\). To confirm these algebraically, set the equation equal to zero: \(x^4 - 2x^3 = 0\). Solving by factoring, you get \(x^3(x - 2) = 0\), resulting in solutions \(x = 0\) and \(x = 2\). This verifies that the x-intercepts matched the visual findings.
Y-Intercept
To find the y-intercept of a polynomial equation, look for the point where the graph crosses the y-axis, which is where \(x = 0\). This gives us a straightforward way to find the intercept: simply substitute \(x = 0\) into the equation. For our equation, \(y = x^4 - 2x^3\), substituting \(x = 0\) simplifies to \(y = 0^4 - 2 \times 0^3 = 0\). Thus, the y-intercept is at \((0, 0)\).

It's essential to compare this with the graph from your calculator – they should align. The intersection at \((0, 0)\) signifies the point where both intercepts coincide, which often highlights a crucial point in polynomial equations, indicating shared roots or solutions.
Factoring Polynomial Equations
Factoring is a robust method for solving polynomial equations and finding intercepts. It involves expressing the polynomial as a product of its simpler polynomial factors. For the equation \(x^4 - 2x^3 = 0\), first observe common terms that can be factored. Here, \(x^3\) is a common factor: \(x^3(x - 2) = 0\).

Once factored, solve each factor set to zero:
  • \(x^3 = 0\) gives \(x = 0\)
  • \(x - 2 = 0\) gives \(x = 2\)
This provides the x-intercepts \((0, 0)\) and \((2, 0)\), just as found graphically and confirmed algebraically. Factoring not only confirms solutions but also simplifies solving higher-degree polynomial equations, breaking them down into manageable parts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Falling Ball Using calculus, it can be shown that if a ball is thrown upward with an initial velocity of \(16 \mathrm{ft} / \mathrm{s}\) from the top of a building 128 ft high, then its height \(h\) above the ground \(t\) seconds later will be $$ h=128+16 t-16 t^{2} $$ During what time interval will the ball be at least \(32 \mathrm{ft}\) above the ground? (image cannot copy)

Without using a calculator, determine which number is larger in each pair. (a) \(2^{1 / 2}\) or \(2^{1 / 3}\) (b) \(\left(\frac{1}{2}\right)^{1 / 2}\) or \(\left(\frac{1}{2}\right)^{1 / 3}\) (c) \(7^{1 / 4}\) or \(4^{1 / 3}\) (d) \(\sqrt[3]{5}\) or \(\sqrt{3}\)

Distances in a City \(A\) city has streets that run north and south and avenues that run east and west, all equally spaced. Streets and avenues are numbered sequentially, as shown in the figure. The walking distance between points \(A\) and \(B\) is 7 blocks - that is, 3 blocks east and 4 blocks north. To find the straight-line distance \(d\), we must use the Distance Formula. (a) Find the straight-line distance (in blocks) between \(A\) and \(B\) (b) Find the walking distance and the straight-line distance between the corner of 4 th St. and 2 nd Ave. and the corner of 11 th St. and 26 th Ave. (c) What must be true about the points \(P\) and \(Q\) if the walking distance between \(P\) and \(Q\) equals the straight[line distance between \(P\) and \(Q ?\)

Sketch the region given by the set. $$\left\\{(x, y) | x^{2}+y^{2} \leq 1\right\\}$$

DISCOVER - PROVE: Relationship Between Solutions and Coefficients The Quadratic Formula gives us the solutions of a quadratic equation from its coefficients. We can also obtain the coefficients from the solutions. (a) Find the solutions of the equation \(x^{2}-9 x+20=0\) and show that the product of the solutions is the constant term 20 and the sum of the solutions is \(9,\) the negative of the coefficient of \(x\) (b) Show that the same relationship between solutions and coefficients holds for the following equations:$$ \begin{array}{l}x^{2}-2 x-8=0 \\\x^{2}+4 x+2=0\end{array}$$ (c) Use the Quadratic Formula to prove that in general, if the equation \(x^{2}+b x+c=0\) has solutions \(r_{1}\) and \(r_{2}\) then \(c=r_{1} r_{2}\) and \(b=-\left(r_{1}+r_{2}\right)\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.