Chapter 1: Problem 8
Sketch the graph of the function. $$ y=\sqrt{4-x^{2}} \text { (Hint: See Example 7.) } $$
Short Answer
Expert verified
Graph the upper half of a circle centered at the origin with radius 2.
Step by step solution
01
Recognize the Shape of the Graph
The given function is \( y = \sqrt{4 - x^2} \). This resembles the standard equation of a circle, \( x^2 + y^2 = r^2 \). In this modified form, it represents the upper half of a circle centered at the origin with a radius of 2.
02
Determine Domain and Range
To ensure the expression under the square root is non-negative, set \( 4 - x^2 \geq 0 \). Solving this inequality gives \( -2 \leq x \leq 2 \). The range of \( y \) is from 0 to 2, as \( y \) will be non-negative and at its maximum when \( x = 0 \).
03
Sketch the Circle
Graph the full circle with the equation \( x^2 + y^2 = 4 \). This is a circle with center (0,0) and radius 2. However, we only consider the upper half of this circle due to the square root function, which only returns non-negative values.
04
Plot Key Points on the Graph
Identify and plot key points: \((-2,0)\), \((0,2)\), \((2,0)\). These represent the leftmost, topmost, and rightmost points on the semicircle.
05
Draw the Upper Semicircle
Connect the points with a smooth, curve which is part of a circle spanning from \((-2,0)\) to \((2,0)\), representing the upper half of a circle. This is the graph of the function \( y = \sqrt{4 - x^2} \).
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.
Domain and Range
The domain and range of a function are fundamental concepts when sketching graphs. For the function \( y = \sqrt{4 - x^2} \), understanding these concepts helps visualize its shape correctly.
- Domain: The domain refers to all possible input values for \( x \) that keep the expression under the square root non-negative. For the given function, the expression is \( 4 - x^2 \). We solve the inequality \( 4 - x^2 \geq 0 \) to find the domain. Rearranging gives \( -2 \leq x \leq 2 \), meaning \( x \) can be any value between -2 and 2.
- Range: The range of the function consists of all possible output values for \( y \). Since \( y \) represents a square root, it will always be non-negative. Here, \( y \) achieves a maximum value of 2 when \( x = 0 \), resulting in a range of \( 0 \leq y \leq 2 \).
Circle Equations
Circle equations are essential to understanding the shape formed by quadratic equations. The standard form of a circle's equation is \( x^2 + y^2 = r^2 \), where \( r \) is the radius.
In our exercise, \( y = \sqrt{4 - x^2} \) is related to the equation \( x^2 + y^2 = 4 \). This is a circle with:
In our exercise, \( y = \sqrt{4 - x^2} \) is related to the equation \( x^2 + y^2 = 4 \). This is a circle with:
- Center: Located at (0,0), the origin.
- Radius: The radius, \( r \), is 2 since \( 4 \) is the square of the radius.
Semicircle
A semicircle is simply half of a circle. For the function \( y = \sqrt{4 - x^2} \), this specifically refers to the upper half of the circle with equation \( x^2 + y^2 = 4 \).
A semicircle maintains the same:
A semicircle maintains the same:
- Center: The midpoint of the diameter; here, it is (0,0).
- Radius: Stays constant at 2 in this exercise.
- Shape: A smooth curve from one point on the horizontal axis to another, but only covering the top half.