/*! 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 26 y=\sqrt{25-x^{2}},-2 \leq x \leq... [FREE SOLUTION] | 91Ó°ÊÓ

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y=\sqrt{25-x^{2}},-2 \leq x \leq 3

Short Answer

Expert verified
The function describes a semi-circle with radius 5, evaluated within the domain \(-2 \leq x \leq 3\).

Step by step solution

01

Understand the Function

The given function is \( y = \sqrt{25 - x^2} \). This is essentially a semi-circle equation derived from the circle equation \( x^2 + y^2 = 25 \), which describes a circle with a radius of 5 centered at the origin in the coordinate plane.
02

Determine Domain Constraints

The domain constraint given is \(-2 \leq x \leq 3\). This means we will only consider values of \(x\) within this interval. The square root expression \( \sqrt{25 - x^2} \) is defined when \( 25 - x^2 \geq 0 \). Solving \( x^2 \leq 25 \), we get \(-5 \leq x \leq 5\). However, we are only interested in the given interval \(-2 \leq x \leq 3\).
03

Evaluate the Function within the Domain

To find values of \(y\), calculate it for some key values of \(x\) in the interval:- For \(x = -2\), \( y = \sqrt{25 - (-2)^2} = \sqrt{25 - 4} = \sqrt{21} \).- For \(x = 0\), \( y = \sqrt{25 - 0^2} = \sqrt{25} = 5 \).- For \(x = 3\), \( y = \sqrt{25 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \).
04

Describe the Graph of the Function

Since we are dealing with the top half of a circle, the graph is a semi-circle from \(-x\) to \(x\), within the specified domain. The vertex of this semi-circle would be at \( (0, 5) \), which is the topmost point reached when \(x = 0\). Given the evaluated points, the function covers a section of this semi-circle, and it smoothly connects through these points due to the continuous nature of the square root function.

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Key Concepts

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

Domain Constraints
The domain constraints play a vital role in identifying the values of \(x\) for which the function is defined. For the function \( y = \sqrt{25 - x^2} \), the terms under the square root must be non-negative for \(y\) to be real and defined. This means we need \( 25 - x^2 \geq 0 \). Solving this inequality leads us to \( x^2 \leq 25 \), giving us the range of \(-5 \leq x \leq 5\).
However, our problem specifies an additional domain constraint: \(-2 \leq x \leq 3\). Therefore, while \(x\) could be theoretically between \(-5\) and \(5\) for the function to be defined, we only consider this smaller interval. Domain constraints hence "trim down" the possible values of \(x\), simplifying the evaluation of the function to this smaller, specified interval.
Function Evaluation
To evaluate the function and understand what \( y = \sqrt{25 - x^2} \) produces within the given domain, we calculate \(y\) for specific values of \(x\) that lie within the interval\(-2 \leq x \leq 3\). This practical step involves substituting these values into the function to get exact \( y \) values, which helps in plotting the graph and understanding its behavior. A few key evaluations are:
  • When \( x = -2 \), the function yields \( y = \sqrt{21} \).
  • Setting \( x = 0 \), we find \( y = 5 \).
  • For \( x = 3 \), we have \( y = 4 \).
These values illustrate how \( y \) changes from one edge of the domain to the other, showcasing the semicircular shape of the function when graphed. Evaluating at multiple points also ensures that the function's shape maintains continuity, which is an essential property of many geometric functions.
Graph Description
The graph of \( y = \sqrt{25 - x^2} \) within the interval \(-2 \leq x \leq 3\) describes a portion of a semicircle. Specifically, this function represents the upper half of a circle with radius 5, centered at the origin \((0,0)\). The topmost point of this semicircle is \((0, 5)\). This is where the function reaches its maximum value, due to the highest value under the square root when \( x = 0 \).
The graph is smooth and continuous, illustrating the gradual rise and fall of \( y \) as \( x \) moves through the domain. Starting from \( x = -2 \), the graph climbs to reach its peak at \( x = 0 \), and then descends back to a lower \( y \) value at \( x = 3 \), where \( y = 4 \). Such a meticulously defined path reflects the semi-circular arc spanning within the confines of \(x\)'s specified domain.

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