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Chapter 1: Problem 7

Starting with the graph of \(y=\sqrt{x},\) the graph of \(y=1 / x,\) and the graph of \(y=\sqrt{1-x^{2}}\) (the upper unit semicircle), sketch the graph of each of the following functions: $$f(x)=-4+\sqrt{-(x-2)}$$

Short Answer

Expert verified
The graph of \( f(x) = -4 + \sqrt{-(x-2)} \) is a downward shift of the reflection of \( y = \sqrt{x} \), with domain \((-\infty, 2]\).

Step by step solution

01

Identify Base Function

The given function is a transformation of the square root function. The base function is \( y = \sqrt{x} \). We're also given transformations involving negation, horizontal shift, and vertical translation.
02

Analyze Transformations

The function inside the square root, \( -(x-2) \), indicates a reflection across the y-axis due to the negative sign, and a horizontal shift 2 units to the right due to \((x-2)\). The overall function, \( y = -4 + \sqrt{-(x-2)} \), also involves a vertical shift 4 units downward because of the \(-4\).
03

Identify Domain

For the function \( \sqrt{-(x-2)} \) to be defined, \( -(x-2) \geq 0 \) must be true. Solving \(-(x-2) \geq 0\), we find \( x \leq 2 \). Hence, the domain of the function is \((-\infty, 2]\).
04

Sketch the Transformation

Start with the reflection of \( y = \sqrt{x} \), which becomes \( y = \sqrt{-x} \), flipping it over the y-axis. Then perform a horizontal shift 2 units to the right, resulting in \( y = \sqrt{-(x-2)} \). Lastly, apply the vertical shift by moving the graph 4 units downward to get \( y = -4 + \sqrt{-(x-2)} \).

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Key Concepts

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

Reflection Across Y-Axis
When a function reflects across the y-axis, it effectively mirrors or reverses horizontally. This occurs when we replace every \( x \)in the function with \( -x \). Here, with the function \( y = \sqrt{-(x-2)} \), because we have a \( - \) within the square root, this indicates a horizontal reflection.
  • Original function is \( y = \sqrt{x} \)
  • Reflection across the y-axis changes it to \( y = \sqrt{-x} \)
  • The graph of \( y = \sqrt{-x} \) reflects the part of the graph that was on the right side of the y-axis to the left.
This reversal fundamentally changes the appearance of the graph since the values for positive \( x\) are now accessed by negative \( x \) values.
Horizontal Shift
A horizontal shift in a function moves the graph left or right. This occurs when we adjust \( x \)within the function, such as \( y = \sqrt{-(x-2)} \).
  • Horizontal shifts are counterintuitive based on signage: \( (x - c) \) represents a shift to the right by \( c \) units.
  • In this case, \( (x - 2) \) suggests we shift the original function to the right by 2 units.
This shift modifies the range of x-values where the graph appears without changing its shape.
Vertical Transformation
Vertical transformations involve moving a graph up or down on the coordinate plane. In our given function \( y = -4 + \sqrt{-(x-2)} \), we have a downward shift.
  • The \( -4 \) indicates that we move the entire graph of \( \sqrt{-(x-2)} \) down by 4 units.
  • Vertical transformations do not affect the horizontal position of the graph.
Such transformations simply elevate or lower the vertical positioning of the graph along the y-axis, altering the function's values without changing its horizontal stretch.
Domain Identification
Determining the domain of a function involves finding all possible values of \( x \) for which the function is defined. For the function \( y = \sqrt{-(x-2)} \), we must ensure the expression inside the square root remains non-negative, since square roots of negative numbers are not defined in the set of real numbers.
  • Thus, we solve \( -(x-2) \geq 0 \), simplifying to \( x \leq 2 \).
  • This defines the domain as all \( x \leq 2 \), or in interval notation, \((-\infty, 2]\).
By understanding the domain, we define where the graph exists on the x-axis. This step is crucial for correctly sketching and interpreting function behaviors.

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Most popular questions from this chapter

Find the standard equation of the circle passing through (-2,1) and tangent to the line \(3 x-2 y=6\) at the point (4,3) . Sketch. (Hint: The line through the center of the circle and the point of tangency is perpendicular to the tangent line.) \(\Rightarrow\)

Starting with the graph of \(y=\sqrt{x},\) the graph of \(y=1 / x,\) and the graph of \(y=\sqrt{1-x^{2}}\) (the upper unit semicircle), sketch the graph of each of the following functions: $$y=f(x)=x /(1-x)$$

Find the domain of each of the following functions: $$y=f(x)=\sqrt[4]{x} \Rightarrow$$

Find the domain of each of the following functions: $$y=f(x)=\sqrt{1-(1 / x)}=$$

Starting with the graph of \(y=\sqrt{x},\) the graph of \(y=1 / x,\) and the graph of \(y=\sqrt{1-x^{2}}\) (the upper unit semicircle), sketch the graph of each of the following functions: $$f(x)=\sqrt{x-2}$$
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