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Chapter 2: Problem 26

Describe in words how the graph of the given function can be obtained from the graph of \(y=x^{2}\) by rigid or nonrigid transformations. $$ f(x)=(x+6)^{2} $$

Short Answer

Expert verified
Shift the graph of \(y = x^2\) 6 units to the left.

Step by step solution

01

Understand the parent function

The parent function here is a basic parabola, given by \(y = x^2\). This is a U-shaped graph centered at the origin, (0,0).
02

Identify the transformation

In the given function, \(f(x) = (x + 6)^2\), notice the expression \((x + 6)\). This indicates a horizontal shift.
03

Determine the direction and magnitude of the shift

The function \((x + 6)^2\) means that the graph of \(y = x^2\) is shifted to the left by 6 units. This is because adding 6 inside the function \((x + 6)\) shifts it in the negative direction on the x-axis.
04

Describe the transformation

Thus, the graph of \(f(x) = (x + 6)^2\) is obtained from the graph of \(y = x^2\) by shifting it 6 units to the left along the x-axis. This is a rigid transformation, as the shape of the graph does not change, only its position.

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Key Concepts

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

Parabola
A parabola is a unique curve on a graph that looks like a U or an upside-down U, formally known as a quadratic function. It has a standard mathematical expression in the form of \( y = ax^2 + bx + c \). When observed on a graph, a simple parabola such as \( y = x^2 \) has its vertex, or the lowest point, at the origin \( (0,0) \). The parabola's symmetry makes it a central topic in algebra and pre-calculus.
  • The vertex of a basic parabola \( y = x^2 \) is the point of reflection, where the graph is perfectly symmetric.
  • The axis of symmetry runs vertically through the vertex, keeping the parabolic shape even on both sides.
  • Changing the coefficients \( a \), \( b \), and \( c \) in the quadratic expression modifies the width, orientation, and position of the parabola.
Understanding the simple structure of a parabola is crucial when learning about function transformations, as it serves as the starting point in many graphing scenarios.
Horizontal Shift
A horizontal shift is a transformation that moves a graph left or right along the x-axis. When functions have changes inside the parentheses (like \(f(x) = (x + h)^2\)), it indicates a horizontal shift.
  • If you add a positive number inside the parenthesis, like \(+6\) in \((x+6)^2\), the graph shifts to the left that number of units.
  • If you subtract a positive number, like \(-6\) in \((x-6)^2\), the graph moves right that many units.
This is counterintuitive for many students, as positive moves left on the x-axis and negative moves right. What happens on the inside (inside the parentheses) is the opposite of what's intuitive for the horizontal movement in a graph.
Rigid Transformation
Rigid transformations involve moving a graph without altering its shape or size. The essential characteristics of rigid transformations are translation (shifting), reflection, and rotation. In the context of a simple quadratic function like \( y = x^2 \), moving the entire curve to a new position is a rigid transformation.
  • Translation changes the location of the graph. For example, \( f(x) = (x+6)^2 \) represents a translation of 6 units to the left.
  • Reflection flips the graph over a specific line, usually the x-axis or y-axis.
  • Rotation spins the graph around a point, though it's less common in simple parabola transformations.
For the given problem \( f(x) = (x+6)^2 \), the rigid transformation is purely translational, maintaining the parabola's original U-shape while shifting it horizontally.

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Most popular questions from this chapter

Discuss how to graph the function \(y=|x|+\mid x-\) 3|. Carry out your ideas.

Suppose \(f\) is a continuous function that is increasing (or decreasing) for all \(x\) in its domain. Explain why \(f\) is necessarily one-to-one.

Proceed as in Example 1 and translate the words into an appropriate function. Give the domain of the function. Express the area of an equilateral triangle as a function of the length \(s\) of one of its sides.

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The given function \(f\) is one-toone. The domain and range of \(f\) is given. Find \(f^{-1}\) and give its domain and range. $$ f(x)=2+\frac{3}{\sqrt{x}}, \quad x>0, y>2 $$
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