Chapter 2: Problem 92
Sketch the graphs of (a) \(y=|x|\) (b) \(y=\frac{1}{2}(x+|x|)\) (c) \(y=|x+1|\) (d) \(y=|x|+|x+1|-2|x+2|+3\) (e) \(|x+y|=1\)
Short Answer
Expert verified
Sketch V, flat and rising lines, shifted V, piecewise segments, and a diamond shape.
Step by step solution
01
Understanding the Graph of y=|x|
The graph of the function \(y=|x|\) is a V-shaped graph. The absolute value function reflects negative input values over the x-axis. Thus, for all positive values of \(x\), \(y\) equals \(x\). For all negative values, \(y\) equals \(-x\). Plot points like \((-1,1)\), \((0,0)\), and \((1,1)\) and draw lines forming a V.
02
Understanding the Graph of y=\frac{1}{2}(x+|x|)
For \(x \geq 0\), \(y = \frac{1}{2}(x+x) = x\). For \(x < 0\), \(y = \frac{1}{2}(x-x) = 0\). Thus, the graph is a line \(y=x\) for \(x \geq 0\), and it is flat along \(y=0\) for \(x<0\). The graph has a corner at the origin \((0,0)\).
03
Understanding the Graph of y=|x+1|
This is a horizontal shift of the \(y=|x|\) graph. The vertex is at \((-1,0)\). The characteristic V-shape starts from \((-1,0)\), rising in both directions. For \(x \geq -1\), \(y = x+1\), and for \(x < -1\), \(y = -(x+1) = -x -1\).
04
Understanding the Graph of y=|x|+|x+1|-2|x+2|+3
This requires examining each piecewise segment. For simplicity, focus segments :1. For \(x < -2\), \(y= -x-x-1-2x-4+3 = -4x-2\).2. For \(-2 \leq x < -1\), \(y= -x-x-1+0+3 = -2x+2\).3. For \(-1 \leq x < 0\), \(y= -x+0+2x+4+3 = x+3\).4. For \(x \geq 0\), \(y= x+x+1-2x-4+3 = 0\). Graph each segment carefully.
05
Understanding the Graph of |x+y|=1
This represents a diamond shape centered at the origin. Rearrange to \( |x+y| = 1 \) translates to four lines: 1. For \(x+y = 1\). 2. For \(x+y = -1\). A square rotated 45 degrees with vertices at \((1,0)\), \((0,1)\), \((-1,0)\), and \((0,-1)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Value Functions
The absolute value function, like the one expressed as \(y=|x|\), is a common and foundational type of mathematical graph. It is characterized by its distinctive V-shaped structure. This V-shape occurs because the absolute value function takes any negative input and reflects it across the x-axis into a positive output. Mathematically, this means for any non-negative x-values, \(y = x\); and for negative x-values, \(y = -x\). Thus, it results in a graph where the vertex or lowest point is at the origin, for the function \(y=|x|\). To quickly sketch this function, you can plot key points such as \((-1,1)\), \((0,0)\), and \((1,1)\) and draw straight lines connecting them, forming the characteristic V shape.
Piecewise Functions
Piecewise functions are composed of different expressions, depending on the input value (x) range. A classic example is the function \(y = \frac{1}{2}(x + |x|)\). Here, the function behaves differently for positive and negative x-values. Specifically, when \(x \geq 0\), you have \(y = x\) because \(|x| = x\); creating a line with a 45-degree slope through the origin. However, when \(x < 0\), the function evaluates to \(y = 0\), as the terms cancel each other out. This results in a flat horizontal line for negative x-values. The overall graph here is linear for positive x and horizontal for negative x, with a sharp turn or corner at the origin.
Horizontal and Vertical Shifts
Graphs of functions can undergo transformations such as horizontal and vertical shifts. Consider the absolute value function \(y=|x+1|\). This involves a horizontal shift of the standard absolute value function, \(y=|x|\). Here, the horizontal shift moves the entire V-shape one unit to the left. The vertex of the V is thus at the point \((-1,0)\). After shifting, the graph's basic structure remains unchanged, maintaining the same slope. The general rule for horizontal shifts involves changing \(x\) to \(x+c\), shifting the graph by \(-c\) units. Vertically, a function would shift up or down based on an added value, although this example specifically concerns horizontal shifts.
Diamond-shaped Graphs
A diamond-shaped graph can emerge from an absolute value equation involving two variables, like \(|x+y|=1\). This equation models a square rotated 45 degrees, centered at the origin. Rearranging the equation leads to four linear relationships: \(x+y=1\) and \(x+y=-1\), essentially linking their intercepts to four points:\((1,0)\), \((0,1)\), \((-1,0)\), and \((0,-1)\). Joining these vertices forms the diamond. This distinct form appears when the sum of absolute values of two linear expressions equals a constant, visually contrasting against lesser-known graph shapes derived from equations. The diamond's center aligns with the cross of the coordinate axes, providing a symmetrical and visually intuitive graph.