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Chapter 0: Problem 26

Finding Intercepts In Exercises \(17-26,\) find any intercepts. $$ y=2 x-\sqrt{x^{2}+1} $$

Short Answer

Expert verified
The y-intercept is at (0,-1) and the x-intercepts are at (\(\sqrt{1/3}\), 0) and (-\(\sqrt{1/3}\), 0).

Step by step solution

01

Find the Y-Intercept

To find the y-intercept, set \(x = 0\) in the given equation and solve for \(y\). \[y = 2(0) - \sqrt{(0)^{2} + 1}\] This simplifies to \[y = -1\] So, the y-intercept is (0,-1).
02

Find the X-Intercept(s)

To find the x-intercepts, set \(y = 0\) in the equation and solve for \(x\). We have \[0 = 2x - \sqrt{x^{2} + 1}\] Moving the square root term to the other side gives \[2x = \sqrt{x^{2} + 1}\] Squaring both sides to eliminate the square root and simplifying gives a quadratic equation \[4x^{2} - x^{2} - 1 = 0\] Further simplifying gives \[3x^{2} - 1 = 0\] Solving this final equation gives \(x = \)卤\(\sqrt{1/3}\), which are the x-intercepts.

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Key Concepts

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

Understanding the Y-Intercept
The y-intercept is the point where a graph crosses the y-axis. At this point, the value of x is always zero because it is the starting line on the y-axis. To find the y-intercept of an equation, substitute 0 for x and solve for y.
For example, in the equation \(y = 2x - \sqrt{x^{2} + 1}\), substitute 0 for \(x\):
  • \(y = 2(0) - \sqrt{(0)^{2} + 1}\)
  • After simplifying, we get \(y = -1\).
So, the y-intercept is the point \((0, -1)\), meaning the graph will intersect the y-axis at -1.
Understanding the X-Intercepts
The x-intercepts are the points where a graph crosses the x-axis. At these points, the value of y is zero because the graph touches or moves through the x-axis.
To find the x-intercepts of an equation, set y to zero and solve for x.
For the equation \(y = 2x - \sqrt{x^{2} + 1}\), substitute 0 for \(y\):
  • \(0 = 2x - \sqrt{x^{2} + 1}\)
  • Move the square root term to the other side: \(2x = \sqrt{x^{2} + 1}\)
  • Square both sides to eliminate the square root: \((2x)^{2} = x^{2} + 1\)
  • Simplify to obtain a quadratic equation: \(3x^{2} - 1 = 0\)
  • Solving this equation results in two x-intercepts: \(x = \pm\sqrt{1/3}\)
These x-intercepts indicate where the graph crosses the x-axis.
What is a Quadratic Equation?
A quadratic equation is a polynomial equation of the form \(ax^{2} + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\).
Quadratic equations often have two solutions, due to the \(x^{2}\) term which creates a parabolic curve when graphed.
To solve quadratic equations, various methods can be employed, such as:
  • Factoring
  • Completing the square
  • Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
Quadratic equations can sometimes be hidden in more complex forms, like our equation \(4x^{2} = x^{2} + 1\), which simplifies to \(3x^{2} - 1 = 0\).
Understanding how to manipulate and solve these is essential for analyzing the intercepts in equations.
The Process of Solving Equations
Solving equations is the mathematical process of finding the values of variables that make an equation true.
There are various methods to solve equations, such as substituting, balancing, factoring, applying the quadratic formula, and more.
Here鈥檚 how you solve the equation \(3x^{2} - 1 = 0\):
  • Add 1 to each side: \(3x^{2} = 1\)
  • Divide each side by 3: \(x^{2} = \frac{1}{3}\)
  • Take the square root of both sides: \(x = \pm\sqrt{\frac{1}{3}}\)
Solving step-by-step helps find solutions to where the graph intercepts the axes. It is crucial to practice these techniques to gain confidence and skill in algebra.

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Most popular questions from this chapter

True or False? In Exercises \(93-96,\) determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If a line contains points in both the first and third quadrants, then its slope must be positive.

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Car Performance The time \(t\) (in seconds) required to attain a speed of \(s\) miles per hour from a standing start for a Volkswagen Passat is shown in the table. (Source: Car \& Driver) $$ \begin{array}{|c|c|c|c|c|c|c|c|}\hline s & {30} & {40} & {50} & {60} & {70} & {80} & {90} \\ \hline t & {2.7} & {3.8} & {4.9} & {6.3} & {8.0} & {9.9} & {12.2} \\ \hline\end{array} $$ (a) Use the regression capabilities of a graphing utility to find a quadratic model for the data. (b) Use a graphing utility to plot the data and graph the model. (c) Use the graph in part (b) to state why the model is not appropriate for determining the times required to attain speeds of less than 20 miles per hour. (d) Because the test began from a standing start, add the point \((0,0)\) to the data. Fit a quadratic model to the revised data and graph the new model. (e) Does the quadratic model in part (d) more accurately model the behavior of the car? Explain.

Let \(R\) be the region consisting of the points \((x, y)\) of the Cartesian plane satisfying both \(|x|-|y| \leq 1\) and \(|y| \leq 1 .\) Sketch the region \(R\) and find its area.
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