/*! 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 28 Finding Intercepts In Exercises ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 0: Problem 28

Finding Intercepts In Exercises \(19-28,\) find any intercepts. \(y=2 x-\sqrt{x^{2}+1}\)

Short Answer

Expert verified
The x-intercepts of the function \(y = 2x - \sqrt{x^{2} + 1}\) are at the points \(x = \sqrt{\frac{1}{3}}\) and \(x = -\sqrt{\frac{1}{3}}\). The y-intercept is at the point \(y = -1\).

Step by step solution

01

Finding the X-Intercept

To find the x-intercept, set \(y = 0\) in the given equation and solve for \(x\). This gives: 0 = \(2x - \sqrt{x^{2} + 1}\).
02

Solving for X

Squaring both sides of the equation and simplifying results in: \(x^{2} = \frac{1}{3}\). The solutions are \(x = \sqrt{\frac{1}{3}}\) and \(x = -\sqrt{\frac{1}{3}}\).
03

Finding the Y-Intercept

To find the y-intercept, set \(x = 0\) in the given equation and solve for \(y\). This gives: \(y = 2(0) - \sqrt{(0)^{2} + 1}\) which simplifies to \(y = -1\).

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 91影视!

Key Concepts

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

Understanding the X-Intercept
Finding the x-intercept is an essential part of understanding the graph of an equation. An x-intercept is a point where the graph crosses the x-axis. This means the y-value is 0 at these points. To find the x-intercept, you simply set the equation equal to zero and solve for the variable x.
  • Start by taking the equation you have and replacing y with 0.
  • In the example, taking the equation, \[ y = 2x - \sqrt{x^2 + 1} \] gives you \[ 0 = 2x - \sqrt{x^2 + 1} \]
  • Now, the task is to isolate x by performing mathematical operations to simplify the equation.
The x-intercepts are found where the balance of the equation holds true, meaning those values of x make the equation equal zero. Identifying these points can help visualize where the function reaches or crosses the x-axis on a graph.
Finding the Y-Intercept
The y-intercept is another fundamental concept when analyzing equations. It is the point where the graph crosses the y-axis, meaning the x-value is zero at this intersection. Finding the y-intercept involves a simple substitution.
  • Set x equal to zero in your equation to find the y value when the graph hits the y-axis.
  • For example, take the equation \[ y = 2x - \sqrt{x^2 + 1} \] and substitute x with 0, resulting in: \[ y = 2(0) - \sqrt{(0)^2 + 1} \]
  • This simplifies to \( y = -1 \), making (0, -1) the y-intercept.
Knowing the location of the y-intercept helps in sketching graphs and understanding the behavior of the equation as it starts from a specific point on the y-axis.
Delving into Solving Equations
Solving equations effectively requires understanding how to manipulate mathematical expressions to isolate the variable of interest. When faced with an equation like \[ 0 = 2x - \sqrt{x^2 + 1} \], the goal is to find values of x that make the equation true. Here are the steps:
  • Rearrange the equation to isolate the terms involving x.
  • If necessary, apply algebraic operations like addition, subtraction, multiplication, division, or even squaring both sides to simplify.
  • In the example, squaring helps remove the square root: \[ 0^2 = (2x - \sqrt{x^2 + 1})^2 \] leads to an equation easier to solve for x.
Solving equations means finding the x-values that make the expression equal, often leading to points on graphs that provide insights into the function's behavior.
Explaining Squaring Equations
Squaring equations is a useful technique when solving more complex equations involving square roots. It involves squaring both sides of the equation to eliminate square roots and simplify the mathematical expression.For instance, starting from \[ 0 = 2x - \sqrt{x^2 + 1} \], squaring both sides gives \[ 0^2 = (2x - \sqrt{x^2 + 1})^2 \]. This step is crucial because it transforms the equation into a quadratic form.Here鈥檚 what to remember:
  • Squaring helps in eliminating roots and making the equation easier to solve for x.
  • After squaring, simplify further by expanding and rearranging terms till you isolate x.
  • Be mindful that squaring can sometimes introduce extraneous solutions, so always verify by substituting back into the original equation.
This approach helps when faced with roots, opening a straightforward route to isolate and solve for variables efficiently.

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

Finding Composite Functions In Exercises \(63-66,\) find the composite functions \(f^{\circ} g\) and \(g \circ f\) . Find the domain of each composite function. Are the two composite functions equal? $$\begin{array}{l}{f(x)=\frac{1}{x}} \\ {g(x)=\sqrt{x+2}}\end{array}$$

Rate of Change In Exercises 63 and \(64,\) you are given the dollar value of a product in 2016 and the rate at which the value of the product is expected to change during the next 5 years. Write a linear equation that gives the dollar value \(V\) of the product in terms of the year t. Let \(t=0\) represent \(2010 .\)) $$\begin{array}{ll}{2016 \text { Value }} & {\text { Rate }} \\ {\$ 1850} & {\$ 250 \text { increase per year }}\end{array}$$

Sketching a Graph of a Function In Exercises \(31-38,\) sketch a graph of the function and find its domain and range. Use a graphing utility to verify your graph. \(f(x)=x+\sqrt{4-x^{2}}\)

Think About It How do the ranges of the cosine function and the secant function compare?

Think About It Sketch the graphs of $$f(x)=\sin x, \quad g(x)=|\sin x|, \quad \text { and } \quad h(x)=\sin (|x|)$$ In general, how are the graphs of \(|f(x)|\) and \(f(|x|)\) related to the graph of \(f ?\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation
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.