Chapter 7: Problem 15
Graph each function. State the domain and range of each function. \(y=-\sqrt{2 x+1}\)
Short Answer
Expert verified
Domain: \(x \geq -\frac{1}{2}\);
Range: \(y \leq 0\).
Step by step solution
01
Recognize the Function Type
The function given is in the form \(y = -\sqrt{2x+1}\), which is a square root function with a negative sign outside the square root, indicating a reflection downwards across the x-axis.
02
Determine the Domain
The expression under the square root \(2x + 1\) must be greater than or equal to zero for the function to be defined. Solve the inequality:\[2x + 1 \geq 0\]Subtract 1 from both sides:\[2x \geq -1\]Divide by 2:\[x \geq -\frac{1}{2}\]Thus, the domain of the function is \(x \geq -\frac{1}{2}\).
03
Determine the Range
Since the square root component \(\sqrt{2x+1}\) produces non-negative outputs and is multiplied by -1, \(y\) will be non-positive. As \(x\) approaches infinity, \(y\) can approach negative infinity. The largest value \(y\) can take (occurring at the boundary \(x = -\frac{1}{2}\)) is zero. Thus, the range of the function is \(y \leq 0\).
04
Plot Key Points
To graph the function, calculate some specific points. Start with the boundary point:- For \(x = -\frac{1}{2}\), \(y = -\sqrt{2(-\frac{1}{2})+1} = 0\).Choose a few values of \(x\) greater than \(-\frac{1}{2}\):- For \(x = 0\), \(y = -\sqrt{2(0)+1} = -1\).- For \(x = 1\), \(y = -\sqrt{2(1)+1} = -\sqrt{3} \approx -1.732\).Plot these points on a coordinate plane: \((-\frac{1}{2}, 0)\), \((0, -1)\), \((1, -1.732)\).
05
Draw the Graph
Sketch a smooth curve that begins at \((-\frac{1}{2}, 0)\) and passes through the calculated points. The curve should continue downwards towards negative infinity as \(x\) increases, reflecting the constraints of the domain and range.
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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 provide crucial information on where the function exists and what values it can take. Understanding these concepts for a square root function helps shed light on the nature of these mathematical curves.
- Domain: The domain of a function includes all the possible input values (x-values) where the function is defined. For the function \(y = -\sqrt{2x + 1}\), we are working with a square root function, which means the expression under the square root must be non-negative. This is because the square root of a negative number is not real. Hence, we solve the inequality \(2x + 1 \geq 0\) to find the x-values that keep the function defined. Solving gives \(x \geq -\frac{1}{2}\). This means the function can accept any x-value greater than or equal to \(-\frac{1}{2}\).
- Range: The range represents all the possible output values (y-values) of the function. Since the function has a negative square root, the values of \(y\) are inverted. Normally, a square root is non-negative, but multiplied by \(-1\), the outputs are all non-positive. At \(x = -\frac{1}{2}\), the value reaches its maximum of 0. As \(x\) grows larger, the outputs become more negative, moving toward negative infinity. Consequently, the range is \(y \leq 0\).
Square Root Function
A square root function exhibits a distinct curve that starts at a specific point and continues indefinitely. For the function \(y = -\sqrt{2x + 1}\), understanding its behavior is key in graphing and interpreting it.
The basic form \(y = \sqrt{x}\) shows a gentle upward curve as x increases. In this exercise, we deal with \(\sqrt{2x + 1}\). The expression under the root, \(2x+1\), shifts the starting point of the basic square root curve to the left by \(\frac{1}{2}\) on the x-axis. Hence, it starts at \(x = -\frac{1}{2}\).
The square root component, \(\sqrt{2x+1}\), produces values from 0 upwards for increasing x-values. It means without any modification to the function, it would yield non-negative y-values.
With the negative sign in front of the square root, the output is reversed, leading to its reflection across the x-axis, which is explained further.
The basic form \(y = \sqrt{x}\) shows a gentle upward curve as x increases. In this exercise, we deal with \(\sqrt{2x + 1}\). The expression under the root, \(2x+1\), shifts the starting point of the basic square root curve to the left by \(\frac{1}{2}\) on the x-axis. Hence, it starts at \(x = -\frac{1}{2}\).
The square root component, \(\sqrt{2x+1}\), produces values from 0 upwards for increasing x-values. It means without any modification to the function, it would yield non-negative y-values.
With the negative sign in front of the square root, the output is reversed, leading to its reflection across the x-axis, which is explained further.
Reflection Across the x-axis
Reflections are transformations that flip a function over a specific axis. In our function \(y = -\sqrt{2x + 1}\), we observe a reflection across the x-axis.
- Understanding Reflection: When a function is multiplied by \(-1\), its graph is flipped over the x-axis. Every point \((x, y)\) becomes \((x, -y)\). For square root functions, this reflection implies an inversion of all positive outputs to negative values.
- Inverted Curve Example: Normally, \(\sqrt{2x+1}\) gives non-negative outputs like 0, 1, and more, reaching towards infinity as \(x\) increases. With the negative sign out front, these outputs are reversed, showing as 0, -1, and so forth, dropping towards negative infinity.
This reflection alters the function's behavior from moving upwards to swooping downwards. Since we see the function starts at the point \((-\frac{1}{2}, 0)\) and moves indefinitely downwards as \(x\) increases, its reflection across the x-axis emphasizes its unique downward trajectory.