/*! 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 69 Graph \(y=\frac{10}{x^{2}}\) and... [FREE SOLUTION] | 91Ó°ÊÓ

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Graph \(y=\frac{10}{x^{2}}\) and \(y=\frac{-10}{x^{2}}\) on the same set of axes. What relationship exists between the two graphs?

Short Answer

Expert verified
The graphs reflect each other across the x-axis.

Step by step solution

01

Understand the Equations

The two functions given are \( y = \frac{10}{x^2} \) and \( y = \frac{-10}{x^2} \). Both equations have the same denominator, \( x^2 \), and differ only by the sign in the numerator.
02

Determine the Symmetry

The function \( y = \frac{10}{x^2} \) is always positive because both \( 10 \) and \( x^2 \) are positive. It is also an even function, meaning it has y-axis symmetry. Similarly, \( y = \frac{-10}{x^2} \) is always negative and is also an even function with y-axis symmetry.
03

Identify Key Points

Let's identify some key values for \( x \) such as \( x = -2, -1, 1, 2 \) to understand the behavior of both graphs around key points. For \( y = \frac{10}{x^2} \): \( y \) at \( x = 1 \) is 10, at \( x = 2 \) is 2.5. Similarly, for \( y = \frac{-10}{x^2} \), \( y \) at \( x = 1 \) is -10, at \( x = 2 \) is -2.5.
04

Analyze the Behavior as x Approaches Zero

Both functions have asymptotes at x=0. As \( x \) approaches zero, \( y \) for \( y = \frac{10}{x^2} \) moves upwards to infinity, and \( y \) for \( y = \frac{-10}{x^2} \) moves downwards to negative infinity. This indicates vertical asymptotes at \( x = 0 \).
05

Plot the Graph

Plot each function on the same axes: \( y = \frac{10}{x^2} \) is a hyperbola in the first and second quadrants (above the x-axis), and \( y = \frac{-10}{x^2} \) is a hyperbola in the third and fourth quadrants (below the x-axis). The graphs are reflections of each other over the x-axis.
06

Identify the Relationship

Both functions are symmetric around the y-axis and have opposite signs. Therefore, \( y = \frac{-10}{x^2} \) is the reflection of \( y = \frac{10}{x^2} \) across the x-axis. This indicates a sign inversion, reflecting each point of \( y = \frac{10}{x^2} \) across the x-axis to obtain \( y = \frac{-10}{x^2} \).

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Key Concepts

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

Symmetry
In mathematics, the concept of symmetry is quite powerful, as it can simplify our understanding of functions and their graphs. A function is considered an even function if it displays symmetry about the y-axis. This means if you replace \( x \) with \( -x \) in the function, the equation remains unchanged.

For the functions \( y = \frac{10}{x^2} \) and \( y = \frac{-10}{x^2} \), both display this characteristic. Despite the difference in the sign of the numerator, both functions are even functions.
  • \( y = \frac{10}{(-x)^2} = \frac{10}{x^2} \)
  • \( y = \frac{-10}{(-x)^2} = \frac{-10}{x^2} \)
These equations confirm symmetry concerning the y-axis. Hence, you will observe that the graphs of these functions are mirror images across this axis. Even functions are quite intuitive to analyze since their graphical representation allows us to focus on half of the function to understand the whole.
Hyperbola
A hyperbola is a type of curve that consists of two separate branches. These branches are mirror images of each other.

The graphs of the functions \( y = \frac{10}{x^2} \) and \( y = \frac{-10}{x^2} \) form hyperbolas. In the case of \( y = \frac{10}{x^2} \), the hyperbolic branches appear in the first and second quadrants because they are reflected on the y-axis and lie entirely above the x-axis.

Conversely, \( y = \frac{-10}{x^2} \) is present in the third and fourth quadrants, exhibiting its branches below the x-axis. This is due to the negative numerator, which flips the hyperbola vertically. Two hyperbolas of the same shape, but of opposite vertical orientation, signify that one graph is a reflection of the other across the x-axis. Understanding hyperbolas can greatly aid in visualizing and intuiting how these functions interact through their graphical representations.
Graphing Asymptotes
Asymptotes are lines that a graph approaches but never touches or crosses. They provide insight into the behavior of a function as it grows indefinitely.

For the functions \( y = \frac{10}{x^2} \) and \( y = \frac{-10}{x^2} \), the line \( x = 0 \) serves as a vertical asymptote. The graphs draw nearer to this line as the value of \( x \) approaches zero, shooting towards infinity or negative infinity.
  • In \( y = \frac{10}{x^2} \), as \( x \) approaches zero from either direction, \( y \) increases to infinity.
  • For \( y = \frac{-10}{x^2} \), \( y \) decreases towards negative infinity as \( x \) nears zero.
Knowing where these asymptotes sit is crucial for accurately drawing the curves of hyperbolic functions. They act as invisible guidelines, shaping the core path but never intersecting them. Recognizing asymptotes helps students predict and understand function behavior as values grow extremely large or very small, a critical skill in calculus and higher mathematics.

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Most popular questions from this chapter

The problem of finding the perpendicular bisector of a line segment presents itself often in the study of analytic geometry. As with any problem of writing the equation of a line, you must determine the slope of the line and a point that the line passes through. A perpendicular bisector passes through the midpoint of the line segment and has a slope that is the negative reciprocal of the slope of the line segment. The problem can be solved as follows: Find the perpendicular bisector of the line segment between the points for the following. Write the equation in standard form. (a) \((-1,2)\) and \((3,0)\) Find the perpendicular bisector of the line segment between the points \((1,-2)\) and \((7,8)\). The midpoint of the line segment is \(\left(\frac{1+7}{2}, \frac{-2+8}{2}\right)\) \(=(4,3)\). The slope of the line segment is \(m=\frac{8-(-2)}{7-1}\) \(=\frac{10}{6}=\frac{5}{3}\). Hence the perpendicular bisector will pass through the point \((4,3)\) and have a slope of \(m=-\frac{3}{5}\). $$ \begin{aligned} y-3 &=-\frac{3}{5}(x-4) \\ 5(y-3) &=-3(x-4) \\ 5 y-15 &=-3 x+12 \\ 3 x+5 y &=27 \end{aligned} $$ Thus the equation of the perpendicular bisector of the line segment between the points \((1,-2)\) and \((7,8)\) is \(3 x+5 y=27\). (b) \((6,-10)\) and \((-4,2)\) (c) \((-7,-3)\) and \((5,9)\) (d) \((0,4)\) and \((12,-4)\)

\(2 y=x-2\)

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Contains the point \((2,-4)\) and is parallel to the \(y\) axis

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