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Chapter 1: Problem 51

Graph the function using transformations. $$y=-\sqrt{x-2}$$

Short Answer

Expert verified
The function is derived from \( y = \sqrt{x} \), shifted right by 2 units, and reflected over the x-axis.

Step by step solution

01

Identify the Base Function

The base function for this problem is the square root function: \( y = \sqrt{x} \). Understanding the base function helps us apply transformations accurately.
02

Horizontal Shift

The expression inside the square root is \( x-2 \), which indicates a horizontal shift of the base function to the right by 2 units. This means the graph of \( y = \sqrt{x} \) will be moved 2 units to the right.
03

Vertical Reflection

The negative sign outside the square root, \( -\sqrt{x-2} \), reflects the graph across the x-axis. This transformation inverts the vertical orientation of the graph, flipping it upside down.
04

Graph the Transformed Function

Start by graphing the base function \( y = \sqrt{x} \). Then apply the horizontal shift by moving the graph 2 units to the right. Finally, perform the reflection by flipping the graph upside down across the x-axis. The graph originates from the point (2, 0) and descends as \( x \) increases.

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Key Concepts

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

Square Root Function
The square root function is denoted by \( y = \sqrt{x} \). It's one of the basic building blocks in algebra and graphing because of its unique shape.
  • This function only exists for non-negative values of \( x \) since we can't take the square root of a negative number in real numbers.
  • The graph of this function is a curve starting at the origin (0,0), gently rising and curving to the right. It forms half of a sideways parabola.
  • This graph is defined entirely in the first quadrant of the coordinate system, where both \( x \) and \( y \) values are non-negative.
In transformations, this function often serves as the base from which we translate and reflect to create new functions. These alterations allow for a greater understanding and interpretation of various phenomena. Understanding these foundational features of \( y = \sqrt{x} \) is crucial as you move on to analyze specific transformations like horizontal shifts and vertical reflections.
Horizontal Shift
When the expression inside the square root contains a term like \( x - c \), this indicates a horizontal shift of the function \( y = \sqrt{x} \) by \( c \) units. In the given problem, the function reads \( \sqrt{x-2} \), which means:
  • The entire graph of \( \sqrt{x} \) is shifted to the right by 2 units.
  • The new origin of the square root curve becomes (2, 0) instead of (0, 0).
  • The starting point of the graph is moved due to this transformation, indicating that every point on the original function is moved horizontally.
To visualize this, imagine grasping the entire \( \sqrt{x} \) graph and nudging it over without altering its shape or direction. This alteration changes the domain of the function from all non-negative numbers to everything greater than or equal to 2. This means that the function now begins at \( x = 2 \), ensuring it remains defined only for sensible \( x \) values compatible with taking square roots.
Vertical Reflection
Vertical reflection involves flipping a function graph about the x-axis. This happens when there's a negative sign in front of the function. For \( y = -\sqrt{x-2} \), this transformation is applied as it reflects the standard square root function downward.
  • For every point on the curve, its \( y \)-value changes to its opposite, effectively inverting the curve.
  • Before reflection, the function increases upward as \( x \) increases. After reflection, it falls downward.
  • This transformation negates the \( y \)-values which means every positive value turns negative, creating an upside-down mirror image around the x-axis.
Reflecting the function like this can help visually indicate negative results or contexts where a drop is expected instead of growth. By understanding these alterations, namely the flipping of the graph, you can more effectively predict and graph conditions where relationships decrease rather than increase.

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Most popular questions from this chapter

In Exercises 99 and 100 determine whether each statement is true or false. If an even function has an interval where the function is increasing, then it also has to have an interval where the function is decreasing.

In Exercises \(39-50\), evaluate \(f(g(1))\) and \(g(f(2)),\) if possible. $$f(x)=\left(1-x^{2}\right)^{1 / 2}, \quad g(x)=(x-3)^{1 / 3}$$

In Exercises \(95-98,\) explain the mistake that is made. Most money market accounts pay a higher interest with a higher principal. If the credit union is offering \(2 \%\) on accounts with less than or equal to \(\$ 10,000\) and \(4 \%\) on the additional money over \(\$ 10,000,\) write the interest function \(I(x)\) that represents the interest earned on an account as a function of dollars in the account. Solution: \(I(x)=\left\\{\begin{array}{ll}0.02 x & x \leq 10,000 \\\ 0.02(10,000)+0.04 x & x>10,000\end{array}\right.\) This is incorrect. What mistake was made?

In Exercises \(51-60\), show that \(f(g(x))=x\) and \(g(f(x))=x\). $$f(x)=\frac{1}{x}, \quad g(x)=\frac{1}{x} \text { for } x \neq 0$$

For Exercises \(95-98,\) refer to the following: In calculus, the difference quotient \(\frac{f(x+h)-f(x)}{h}\) of a function \(f\) is used to find the derivative \(f^{\prime}\) of \(f\) by letting \(h\) approach \(0, h \rightarrow 0\). Find the derivatives of \(F(x)=x, G(x)=x^{2},\) and \(H(x)=(F+G)(x)=x+x^{2} .\) What do you observe?
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