Problem 53
Question
a. Graph \(y=\sqrt{x-2}-2\) b. Find the domain and the range. b. At what coordinate point des the graph start? d. What is the relationship of the point at which the graph starts to the domain and the range?
Step-by-Step Solution
Verified Answer
The graph of the function \(y=\sqrt{x-2}-2\) starts at the point (2,-2). The domain of the function is \([2, \infty)\) and the range of the function is \([-2, \infty)\). The point at which the graph starts is the lower limit or boundary for both the domain and the range of the function.
1Step 1: Graph the function
To graph the function \(y =\sqrt {x-2} - 2\), a starting point is to graph the basic function \(y = \sqrt{x}\). This is a half parabola that starts from the origin (0,0) and grows towards the right side of the graph. Then, the function \(y=\sqrt{x-2}\) is obtained from \(y = \sqrt{x}\) by shifting it 2 units to the right. Now this graph starts at (2,0). Lastly, the function \(y=\sqrt{x-2}-2\) is then obtained from \(y=\sqrt{x-2}\) by shifting it down by 2 units. Now, this graph starts from the point (2,-2).
2Step 2: Find the domain and the range
The domain of a function includes all the possible values of \(x\) for the function. For the given function \(y=\sqrt{x-2}-2\), \(x\) must be greater than or equal to 2, as the square root of a negative number is not a real number. Therefore, the domain is \([2, \infty)\). The range of a function includes all the possible values of \(y\). As the lowest point of the function graph is -2, and it goes upwards from there, the range is \([-2, \infty)\).
3Step 3: Identify the starting point of the graph and relationship to the domain and range
As per the graph, the starting point of the function \(y=\sqrt{x-2}-2\) is (2,-2). This point is related to the domain and range of the function as it signifies the lower limit for both sets. The domain is \([2, \infty)\) and the range is \([-2, \infty)\), showing that the graph starts from the x-value of 2 and y-value of -2. So, the point at which the graph starts is also the lower limit of the domain and range of this function.
Key Concepts
Understanding Domain and RangeFunction Transformation InsightsPinpointing the Starting Point of the Graph
Understanding Domain and Range
The domain and range are essential components when it comes to graphing a function. They define the scope within which a function can operate. For the square root function given by the equation \( y = \sqrt{x-2} - 2 \), the domain and range tell us two crucial pieces of information: what values \( x \) can take, and what values \( y \) can attain.
**Domain** refers to the set of all possible input values (\( x \) values) for the function. Since we cannot take the square root of a negative number in real numbers, \( x \) must be greater than or equal to 2. This means the domain is \([2, \infty)\).
**Range** presents the set of all possible output values (\( y \) values) from the function. Given the function starts at \( -2 \) and increases upwards, the range is \([-2, \infty)\).
**Domain** refers to the set of all possible input values (\( x \) values) for the function. Since we cannot take the square root of a negative number in real numbers, \( x \) must be greater than or equal to 2. This means the domain is \([2, \infty)\).
- \( x \geq 2 \)
- Expressed in interval notation as \([2, \infty)\)
**Range** presents the set of all possible output values (\( y \) values) from the function. Given the function starts at \( -2 \) and increases upwards, the range is \([-2, \infty)\).
- Starts from \( -2 \)
- Expressed in interval notation as \([-2, \infty)\)
Function Transformation Insights
Function transformation involves shifting, stretching, or reflecting the graph of a basic function to produce a more complex form. Let's break down how these transformations apply to the function \( y = \sqrt{x-2} - 2 \) based on the original square root function \( y = \sqrt{x} \).
**Shift to the Right**: By replacing \( x \) in \( y = \sqrt{x} \) with \( x-2 \), the graph shifts horizontally to the right by 2 units. This transformation moves the starting point from the origin (0,0) to (2,0).
**Downward Shift**: Next, by subtracting 2 from \( y = \sqrt{x-2} \), the graph shifts vertically downwards by 2 units. This adjusts our graph's starting point to (2,-2).
**Shift to the Right**: By replacing \( x \) in \( y = \sqrt{x} \) with \( x-2 \), the graph shifts horizontally to the right by 2 units. This transformation moves the starting point from the origin (0,0) to (2,0).
**Downward Shift**: Next, by subtracting 2 from \( y = \sqrt{x-2} \), the graph shifts vertically downwards by 2 units. This adjusts our graph's starting point to (2,-2).
- Horizontal Shift: \( x - 2 \) shifts the graph 2 units to the right.
- Vertical Shift: \( -2 \) means the graph moves downward by 2 units.
Pinpointing the Starting Point of the Graph
The starting point of the graph is a key feature when examining square root functions. For the function \( y = \sqrt{x-2} - 2 \), the starting point is the coordinate (2,-2). This is the first point on the graph where the function begins to take effect.
Finding the starting point involves identifying both horizontal and vertical shifts from the base function \( y = \sqrt{x} \).
The horizontal component \( x-2 \) suggests that the minimum x-coordinate is 2, aligning with the domain's starting value. Meanwhile, adjusting the function by \( -2 \) to give \( y = \sqrt{x-2} - 2 \) sets the initial y-coordinate as -2, mirroring the range's starting point.
Finding the starting point involves identifying both horizontal and vertical shifts from the base function \( y = \sqrt{x} \).
The horizontal component \( x-2 \) suggests that the minimum x-coordinate is 2, aligning with the domain's starting value. Meanwhile, adjusting the function by \( -2 \) to give \( y = \sqrt{x-2} - 2 \) sets the initial y-coordinate as -2, mirroring the range's starting point.
- Domain Starting Point: \( x = 2 \)
- Range Starting Point: \( y = -2 \)
Other exercises in this chapter
Problem 52
Simplify each number. $$32^{1.2}$$
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Simplify each radical expression. Use absolute value symbols when needed. $$ \sqrt[2 n]{x^{2 n}} $$
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For each function \(f,\) find \(f^{-1},\) the domain and range of \(f\) and \(f^{-1},\) and determine whether \(f^{-1}\) is a function. $$ f(x)=(x-4)^{2} $$
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Solve. Check for extraneous solutions. \(\sqrt{x+1}+\sqrt{2 x}=\sqrt{5 x+3}\)
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