/*! 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 5 Identify the underlying basic fu... [FREE SOLUTION] | 91Ó°ÊÓ

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Identify the underlying basic function, and use transformations of the basic function to sketch the graph of the given function. $$h(x)=|x-2|$$

Short Answer

Expert verified
h(x)=|x-2| is the absolute value function |x| that has been transformed by shifting 2 units to the right. The graph of the function is a V-shape with the vertex at the point (2,0).

Step by step solution

01

Identify the Basic Function

The basic function can be identified from looking at the equation. The basic function is the absolute value function |x|. Its graph is V-shaped with the vertex at the origin (0,0).
02

Identify the Transformation

The transformation is determined by the value inside the absolute value operation. Since it is |x-2|, it means the sketch of the graph of the basic function |x| has been shifted 2 units to the right on the x-axis.
03

Sketch the Graph

The graph of the function \(h(x)=|x-2|\) is a V-shape just like the basic function |x|, but moved 2 units to the right of the origin. So, the vertex of the graph is at the point (2,0).

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Key Concepts

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

Transformation of Functions
Understanding how basic functions can be modified to create new graphs is crucial in algebra. The process of altering a graph is known as transformation. There are several types of transformations: shifting, stretching, compressing, and reflecting. If we take the absolute value function as an example, we can shift it horizontally or vertically, stretch or compress it by multiplying it by a factor, or flip it over the x-axis or y-axis.

Shifts in Functions

Shifts are perhaps the most straightforward transformations. For instance, if we take the basic absolute value function, denoted as \( f(x) = |x| \), and want to shift it to the right by 2 units, we would write it as \( h(x) = |x - 2| \). This means every point on the graph of \( f(x) \) is moved 2 units to the right to get the graph of \( h(x) \). These transformations preserve the shape of the graph but change its position.
Absolute Value Graphs
The graph of the absolute value function is distinctive for its V-shape. The formal definition of the absolute value function is \( f(x) = |x| \), which equals x when x is positive or zero, and -x when x is negative. The corner or tip of the 'V' is known as the vertex, and for the basic absolute value function, it is located at the origin of the coordinate system.

Plotting Absolute Value Functions

When asked to graph an absolute value function, you should start by plotting the vertex. From there, you can determine the slope of the lines that form the 'V'. In most basic cases, these slopes will be 1 and -1, resulting in two symmetric lines meeting at the vertex. It's important to note that modifications inside the absolute value signs—involving addition or subtraction—result in horizontal shifts, not changes in the slope of the lines.
Graph Shifts
Graph shifts refer to the horizontal and vertical translations of a function's graph on the coordinate plane. When we discuss shifts, we're talking about moving the entire graph without changing its shape.

Understanding Horizontal and Vertical Shifts

For horizontal shifts, adding or subtracting inside the function moves the graph left or right. For example, \( f(x) = |x-2| \) represents a horizontal shift of the absolute value function 2 units to the right. On the other hand, for vertical shifts, we add or subtract outside the function. If we had \( f(x) = |x| + 3 \), it indicates the graph is lifted 3 units up.

In essence, recognizing these shifts helps to quickly sketch graphs of functions without plotting numerous points. When facing a new function, students should first decipher the basic function and then apply the shifts step by step to arrive at the correct graph.

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Most popular questions from this chapter

A ball is thrown directly upward from ground level at time \(t=0\) ( \(t\) is in seconds). At \(t=3,\) the ball reaches its maximum distance from the ground, which is 144 feet. Assume that the distance of the ball from the ground (in feet) at time \(t\) is given by a quadratic function \(d(t) .\) Find an expression for \(d(t)\) in the form \(d(t)=a(t-h)^{2}+k\) by performing the following steps. (a) From the given information, find the values of \(h\) and \(k\) and substitute them into the expression \(d(t)=a(t-h)^{2}+k\) (b) Now find \(a\). To do this, use the fact that at time \(t=0\) the ball is at ground level. This will give you an equation having just \(a\) as a variable. Solve for \(a\) (c) Now, substitute the value you found for \(a\) into the expression you found in part (a). (d) Check your answer. Is (3,144) the vertex of the associated parabola? Does the parabola pass through (0,0)\(?\)

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Name at least two features of a quadratic function that differ from those of a linear function.

In Exercises \(97-100,\) let \(f(t)=-t^{2}\) and \(g(x)=x^{2}-1\). Evaluate \((f \circ f)(-1)\)

A rectangular garden plot is to be enclosed with a fence on three of its sides and an existing wall on the fourth side. There is 45 feet of fencing material available. (a) Write an equation relating the amount of available fencing material to the lengths of the three sides that are to be fenced. (b) Use the equation in part (a) to write an expression for the width of the enclosed region in terms of its length. (c) For each value of the length given in the following table of possible dimensions for the garden plot, fill in the value of the corresponding width. Use your expression from part (b) and compute the resulting area. What do you observe about the area of the enclosed region as the dimensions of the garden plot are varied? $$\begin{array}{ccc}\hline \begin{array}{c}\text { Length } \\\\\text { (feet) }\end{array} & \begin{array}{c}\text { Width } \\\\\text { (feet) }\end{array} & \begin{array}{c}\text { Total Amount of } \\\\\text { Fencing Material (feet) }\end{array} & \begin{array}{c}\text { Area } \\\\\text { (square feet) }\end{array} \\\\\hline 5 & & 45 \\\10 & & 45 \\\15 & & 45 \\\20 & & 45 \\\30 & & 45 \\\k & & 45 \\\& &\end{array}$$ (d) Write an expression for the area of the garden plot in terms of its length. (e) Find the dimensions that will yield a garden plot with an area of 145 square feet.

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