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Chapter 2: Problem 43

In Problems \(43-46,\) describe in words how the graph of the first function is obtained from the graph of the second function using rigid and nonrigid transformations. Carefully graph the first function. $$ y=-1+2 \sqrt{-x+2} ; y=\sqrt{x} $$

Short Answer

Expert verified
Reflect \( y=\sqrt{x} \) over the y-axis, shift right 2 units, stretch vertically by 2, and shift downward by 1.

Step by step solution

01

Understanding the Functions

We need to compare the given functions: the first function is \( y = -1 + 2 \sqrt{-x+2} \) and the second function is \( y = \sqrt{x} \). We will describe how to transform \( y = \sqrt{x} \) into \( y = -1 + 2 \sqrt{-x+2} \) using transformations.
02

Transformation Step 1: Horizontal Reflection and Shift

The expression \( \sqrt{-x+2} \) involves a horizontal reflection and horizontal shift. The negative sign in \(-x\) reflects the function \( y = \sqrt{x} \) across the y-axis. Additionally, \(-x + 2\) can be rewritten as \(-(x - 2)\), indicating a shift 2 units to the right.
03

Transformation Step 2: Vertical Stretch

The coefficient 2 in \( 2\sqrt{-x+2} \) means a vertical stretch by a factor of 2. This affects the steepness of the original graph, effectively doubling each y-value from the graph of \( \sqrt{-x+2} \).
04

Transformation Step 3: Vertical Shift

The term \(-1\) is a vertical shift that moves the entire graph downward by 1 unit. This transformation affects every point on the graph by decreasing each y-value by 1.
05

Graphing the Transformed Function

Begin by drawing the basic shape of \( y = \sqrt{x} \). Reflect it across the y-axis to represent \( \sqrt{-x} \), then shift it 2 units to the right to achieve \( \sqrt{-x+2} \). Apply a vertical stretch by multiplying all y-values by 2 and, finally, shift the entire graph down by 1 unit to complete \( y = -1 + 2\sqrt{-x+2} \) graphically.

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Key Concepts

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

Horizontal Reflection
A horizontal reflection is an exciting transformation. It flips a graph over the y-axis. If you have a function like \( y = \sqrt{x} \), applying a horizontal reflection involves changing the sign inside the square root, resulting in \( y = \sqrt{-x} \). Now, the graph that used to stretch from the origin toward the right will stretch from the origin toward the left instead. Reflecting a graph is like looking at its mirror image along the y-axis.
  • Points on the right side of the original graph move to the left in the reflection.
  • The shapes remain the same, but reversed.
This is what the negative sign does in the square root of our transformed function. It's the starting point for other transformations too!
Transformations build upon each other, and here, this reflection will be followed by a shift.
Vertical Stretch
Vertical stretch alters the height of the graph, making it taller or shorter. It's like zooming in or out vertically.The transformed function, \( y = 2\sqrt{-x+2} \), involves a vertical stretch.This stretch reflects in a change of scale for the y-values.When multiplying a function by a number greater than one, like 2 in this case, every y-value of \( \sqrt{-x+2} \) becomes twice as large.
  • The graph appears steeper.
  • Each point's vertical distance from the x-axis is doubled.
Visualized, it's as if you're pulling each point of the graph further away from the x-axis.This is particularly important in modifying the function to achieve the desired form step by step.
Vertical Shift
The vertical shift is a simple movement of the graph up or down.In the function \( y = -1 + 2\sqrt{-x+2} \), this is represented by the \(-1\) term. It tells us to move the entire graph of \( 2\sqrt{-x+2} \) downward by one unit.This affects every point on the graph equally.So, each y-value decreases by one:
  • It does not change the shape of the graph.
  • Only its vertical position on the coordinate plane shifts.
Understanding this helps in accurately placing the transformed graph in its final position.Without vertical shift, the graph would remain higher on the plane, not achieving the intended transformation.
Graphing Functions
Graphing a function involves sketching it on a set of axes.For our example, we start with the base function \( y = \sqrt{x} \).Transforming this into \( y = -1 + 2\sqrt{-x+2} \) involves applying each transformation in sequence.Follow these steps:1. Reflect the graph horizontally over the y-axis to obtain \( y = \sqrt{-x} \).2. Shift it to the right by 2 units to achieve \( \sqrt{-x+2} \).3. Apply a vertical stretch, making the graph steeper, providing \( 2\sqrt{-x+2} \).4. Finally, shift it down by 1 unit to get the final form \( y = -1 + 2\sqrt{-x+2} \).Each change alters the graph鈥檚 standing:
  • Reflections change the graph's orientation.
  • Stretches adjust its steepness.
  • Shifts move it to the required position.
Combining these steps in the correct order is key to accurate graphing!

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Most popular questions from this chapter

In Problems \(9-34\), sketch the graph of the given piecewise-defined function. Find any \(x\) - and \(y\) intercepts of the graph. Give any numbers at which the function is discontinuous. $$ y=\left\\{\begin{array}{ll} -x, & x \leq 1 \\ -1, & x>1 \end{array}\right. $$

In Problems \(1-4\), find the indicated values of the given piecewise-defined function \(f\). $$ f(x)=\left\\{\begin{array}{ll} \frac{x^{2}-4}{x-2}, & x \neq 2 \\ 4, & x=2 \end{array} \quad f(0), f(2), f(-7)\right. $$

In Problems \(9-34\), sketch the graph of the given piece wise-defined function. Find any \(x\) - and \(y\) intercepts of the graph. Give any numbers at which the function is discontinuous. $$ y=|x+3| $$

Use the appropriate derivatives obtained in Problems 11-26. For the given function, find the point of tangency and slope of the tangent line at the indicated value of \(x .\) Find an equation of the tangent line at that point. $$ f(x)=x+\frac{1}{x}, \quad x=\frac{1}{2} $$

The function \(f(x)=|2 x-4|\) is not one-to-one. How should the domain of \(f\) be restricted so that the new function has an inverse? Find \(f^{-1}\) and give its domain and range. Sketch the graph of \(f\) on the restricted domain and the graph of \(f^{-1}\) on the same coordinate axes.
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