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Chapter 2: Problem 100

DISCUSS: Obtaining Transformations Can the function \(g\) be obtained from \(f\) by transformations? If so, describe the transformations needed. The functions \(f\) and \(g\) are described algebraically as follows: $$ f(x)=(x+2)^{2} \quad g(x)=(x-2)^{2}+5 $$

Short Answer

Expert verified
Yes, shift 4 units right and 5 units up.

Step by step solution

01

Identify Function Components

First, let's identify the transformations by comparing the functions \( f(x) = (x+2)^2 \) and \( g(x) = (x-2)^2 + 5 \).
02

Horizontal Shift

The function \( f(x) \) has \((x + 2)^2\), while \( g(x) \) has \((x - 2)^2\). This indicates a horizontal shift. To transform \( f \) into \( g \), we shift \( f(x) \) 4 units to the right because \((x + 2)\) is transformed to \((x - 2)\).
03

Vertical Shift

Next, compare the constants. There is an additional \(+5\) in \( g(x) = (x-2)^2 + 5 \), which means there is a vertical shift upwards by 5 units.
04

Transformation Summary

The transformations needed to obtain \( g \) from \( f \) are to shift the function 4 units to the right (horizontal shift) and then 5 units upward (vertical shift).

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Key Concepts

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

Horizontal Shift
When we talk about a horizontal shift in the context of function transformations, we're focusing on how a function graph moves along the x-axis. To understand this better, think about the function
  • The original function, \( f(x) = (x+2)^2 \), includes a term \( (x + 2) \). This term suggests that in the "parent" quadratic function, \(x^2\), we shifted 2 units to the left to reach the state of \(x+2\).
  • Now, consider the function \( g(x) = (x-2)^2 + 5 \). Here, the term \( (x - 2) \) indicates a movement in the opposite direction, resulting in a shift 2 units to the right from the standard position of the "base" function \(x^2\). This difference between \( (x+2) \) in \( f(x) \) and \( (x-2) \) in \( g(x) \) signifies that the horizontal shift required to transform \( f \) to \( g \) is 4 units to the right.
This horizontal shift affects every point of the graph of a function by moving it without altering its shape.
Vertical Shift
A vertical shift involves moving the function graph up or down on the y-axis, modifying the output of the function without changing its basic shape. Let's explore the function transformations again:
  • For the initial function \( f(x) = (x+2)^2 \), there is no added constant outside of the squared term, which means it stays centered vertically.

  • However, looking at \( g(x) = (x-2)^2 + 5 \), we notice the \(+ 5\) at the end. This addition means we lift the entire graph of the function upward by 5 units compared to its base form.
The vertical shift adjusts the starting height of the function, effectively raising or lowering it while maintaining its quadratic shape. All while ensuring that each point of the graph moves uniformly along the vertical axis.
Quadratic Functions
Quadratic functions form a fundamental aspect of algebra and geometry, represented by the standard format \( f(x) = ax^2 + bx + c \). They typically graph into a U-shaped curve known as a parabola.
  • In our examples, both functions \( f(x) = (x+2)^2 \) and \( g(x) = (x-2)^2 + 5 \) adhere to this format, where each is a specific case of \( ax^2 \), with \( a = 1 \). This results in parabola shapes.

  • The vertex of a quadratic function is the maximum or minimum point of the parabola, acting as a central reference. Our functions showcase how transformations, like horizontal and vertical shifts, affect this vertex's position.
Understanding transformations allows us to modify these quadratic functions into various desired positions on a graph, making quadratic functions a versatile tool for modeling physical phenomena and solving mathematical problems.

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Most popular questions from this chapter

DISCUSS: Solving an Equation for an Unknown Function In Exercises \(69-72\) of Section 2.7 you were asked to solve equations in which the unknowns are functions. Now that we know about inverses and the identity function (see Exercise 104 ), we can use algebra to solve such equations. For instance, to solve \(f \circ g=h\) for the unknown function \(f,\) we perform the following steps: \(f^{\circ} g=h \quad\) Problem: Solve for \(f\) \(f^{\circ} g^{\circ} g^{-1}=h^{\circ} g^{-1} \quad\) Compose with \(g^{-1}\) on the right \(f \circ I=h \cdot g^{-1} \quad \text { Because } g^{\circ} g^{-1}=I\) \(f=h e g^{-1} \quad\) Because \(f \circ I=f\) So the solution is \(f=h \circ g^{-1} .\) Use this technique to solve the equation \(f \circ g=h\) for the indicated unknown function. (a) Solve for \(f,\) where \(g(x)=2 x+1\) and \(h(x)=4 x^{2}+4 x+7.\) (b) Solve for \(q,\) where \(f(x)=3 x+5\) and \(h(x)=3 x^{2}+3 x+2.\)

A man is running around a circular track that is 200 m in circumference. An observer uses a stopwatch to record the runner's time at the end of each lap, obtaining the data in the following table. (a) What was the man's average speed (rate) between 68 s and 152 s? (b) What was the man's average speed between 263 s and 412 s? (c) Calculate the man's speed for each lap. Is he slowing down, speeding up, or neither? $$\begin{array}{|c|c|} \hline \text { Time (s) } & \text { Distance (m) } \\ \hline 32 & 200 \\ 68 & 400 \\ 108 & 600 \\ 152 & 800 \\ 203 & 1000 \\ 263 & 1200 \\ 335 & 1400 \\ 412 & 1600 \\ \hline \end{array}$$

Graphing Functions Sketch a graph of the function by first making a table of values. $$H(x)=|2 x|$$

A linear function is given. (a) Find the average rate of change of the function between \(x=a\) and \(x=a+h .\) (b) Show that the average rate of change is the same as the slope of the line. $$g(x)=-4 x+2$$

Sketch graphs of the functions \(f(x)=\|x\|, g(x)=\|2 x\|,\) and \(h(x)=\|3 x\|\) on separate graphs. How are the graphs related? If \(n\) is a positive integer, what does a graph of \(k(x)=\|n x\|\) look like?
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