\chapter{Functions and graphs}
\section{Basic functions and simple transformations}
We now study the equations and graphs of some basic functions, together with some of their properties.
The simplest function we can think of is the function whose output is always the same. Such a function is called
a \textit{constant} function, and its equation is \(y=c\), where \(c\) is some fixed number.
The domain of a constant function is \((-\infty,\infty)\), and its range is the single number \(\{c\}\). The graph is a horizontal straight line, as shown in the picture below.
\[\img{U1_10F2.jpg}{}{12em}{}\]
This function is clearly not invertible, as the horizontal line test fails (badly) if we draw a horizontal line on top
of the graph itself. We will not use constant functions in the rest of this unit, but keep in mind what they are
and what their graph looks like.
\subsection{The seven parent functions}
We now study the equations and graphs of some basic functions, together with some of their properties.
\begin{enumerate}
\item \textbf{ Identity function}
This is the function that leaves the input unchanged. So the equation is \(y=x\) and the graph is a straight line through the origin with slope 1.
\[
\begin{array}{ccc}
\begin{array}{c}
\mbox{A few points}\\
\begin{array}{c|c}
x & y \\ \hline -1& -1\\0 & 0\\1&1\end{array}
\end{array}
& \begin{array}{c}
\mbox{Graph}\\
\img{U1_10F1.jpg}{}{12em}{}\end{array} & \begin{array}{ll}
\mbox{Equation} & y=x\\
\mbox{Domain} &(-\infty,\infty)\\
\mbox{Range} & (-\infty,\infty)\\
\mbox{Invertible} & \mbox{Yes}
\end{array}
\end{array}
\]
\item \textbf{Square function}
This is the function that squares its input. So the equation is \(y=x^2\).
This is the simplest example of a \textit{quadratic function}. We will study quadratic functions
in depth in Chapter 2.
The graph is a \textit{parabola} with
vertex at the origin. Since the square of a real number cannot be negative, the range is
\([0,\infty)\).
\[
\begin{array}{ccc}
\begin{array}{c}
\mbox{A few points}\\
\begin{array}{c|c}
x & y \\ \hline -2& 4\\-1 & 1\\0&0\\1&1\\2&4\end{array}
\end{array}
& \begin{array}{c}
\mbox{Graph}\\
\img{U1_10F3.jpg}{}{12em}{}\end{array} & \begin{array}{ll}
\mbox{Equation} & y=x^2\\
\mbox{Domain} &(-\infty,\infty)\\
\mbox{Range} & [0,\infty)\\
\mbox{Invertible} & \mbox{No}
\end{array}
\end{array}
\]
\item \textbf{Square Root function}
This function has equation \(y=\sqrt{x}\). The square root of a negative number is not a real number,
and \(\sqrt{x}\) is positive or zero, so both domain and range are \([0,\infty)\).
\[
\begin{array}{ccc}
\begin{array}{c}
\mbox{A few points}\\
\begin{array}{c|c}
x & y \\ \hline 0& 0\\1 & 1\\4&2\end{array}
\end{array}
& \begin{array}{c}
\mbox{Graph}\\
\img{U1_10F4.jpg}{}{12em}{}\end{array} & \begin{array}{ll}
\mbox{Equation} & y=\sqrt{x}\\
\mbox{Domain} &[0,\infty)\\
\mbox{Range} & [0,\infty)\\
\mbox{Invertible} & \mbox{Yes}
\end{array}
\end{array}
\]
\item \textbf{Cube function}
The equation is \(y=x^3\). The cube of a negative number is negative.
So the range is \((-\infty,\infty)\).
\[
\begin{array}{ccc}
\begin{array}{c}
\mbox{A few points}\\
\begin{array}{c|c}
x & y \\ \hline -1& -1\\0 & 0\\1&1\end{array}
\end{array}
& \begin{array}{c}
\mbox{Graph}\\
\img{U1_10F5.jpg}{}{12em}{}\end{array} & \begin{array}{ll}
\mbox{Equation} & y=x^3\\
\mbox{Domain} &(-\infty,\infty)\\
\mbox{Range} & (-\infty,\infty)\\
\mbox{Invertible} & \mbox{Yes}
\end{array}
\end{array}
\]
\item \textbf{Cube Root function}
The equation is \(y=\sqrt[3]{x}\). The cube root of a negative number is a negative real number,
so both domain and range are \((-\infty,\infty)\).
\[
\begin{array}{ccc}
\begin{array}{c}
\mbox{A few points}\\
\begin{array}{c|c}
x & y \\ \hline -1& -1\\0 & 0\\1&1\end{array}
\end{array}
& \begin{array}{c}
\mbox{Graph}\\
\img{U1_10F6.jpg}{}{12em}{}\end{array} & \begin{array}{ll}
\mbox{Equation} & y=\sqrt[3]{x}\\
\mbox{Domain} &(-\infty,\infty)\\
\mbox{Range} & (-\infty,\infty)\\
\mbox{Invertible} & \mbox{Yes}
\end{array}
\end{array}
\]
\item \textbf{Absolute Value function}
This is the function with equation \(y=|x|\), that leaves positive numbers
unchanged, and turns negative numbers into
positive ones. By definition, \(|x|\) cannot be negative, so the range is \([0,\infty)\).
\[
\begin{array}{ccc}
\begin{array}{c}
\mbox{A few points}\\
\begin{array}{c|c}
x & y \\ \hline -2& 2\\-1 & 1\\0&0 \\1& 1 \\ 2& 2\end{array}
\end{array}
& \begin{array}{c}
\mbox{Graph}\\
\img{U1_10F7.jpg}{}{12em}{}\end{array} & \begin{array}{ll}
\mbox{Equation} & y=|x|\\
\mbox{Domain} &(-\infty,\infty)\\
\mbox{Range} & [0,\infty)\\
\mbox{Invertible} & \mbox{No}
\end{array}
\end{array}
\]
\item \textbf{Reciprocal function}
This function gives the reciprocal of the input. So the equation is
\(y=\displaystyle{\frac{1}{x}}\), and \(x=0\) cannot be used as an input. Also,
\(y=\displaystyle{\frac{1}{x}}\) cannot be zero, so the range will not contain 0.
\[
\begin{array}{ccc}
\begin{array}{c}
\mbox{A few points}\\
\begin{array}{c|c}
x & y \\ \hline -2& -1/2\\-1 & 1\\1&1 \\2& 1/2\end{array}
\end{array}
& \begin{array}{c}
\mbox{Graph}\\
\img{U1_10F8.jpg}{}{12em}{}\end{array} & \begin{array}{ll}
\mbox{Equation} & y=\dfrac{1}{x}\\
\mbox{Domain} &(-\infty,0)\cup(0,\infty)\\
\mbox{Range} & (-\infty,0)\cup(0,\infty)\\
\mbox{Invertible} & \mbox{Yes}
\end{array}
\end{array}
\]
\end{enumerate}
\subsection{Simple transformations}
We often refer to the basic functions we just described as \textit{parent functions}.
We can get many more new functions by some simple changes to either the input
\(x\) or the output \(f(x)\) of a parent function. The simple changes we will use are
called \textit{simple transformations}.
There are three types of simple transformations: shift, reflection,
and stretching/compressing.
Each of these simple transformations can be applied to the input \(x\), or
the output \(f(x)\), resulting in many different transformed functions.
\subsubsection{Shifts}
A shifted graph is obtained when a number is added to or subtracted from the input or output variable.
| Vertical Shift |
|---|
| \(\begin{array}{c}
\mbox{Let \(h\) be a positive number. Then:}\\
\mbox{The graph of \(f(x)+h\) is the graph of \(f(x)\) shifted up by \(h\)} \\
\mbox{The graph of \(f(x)-h\) is the graph of \(f(x)\) shifted down by \(h\)}
\end{array}\) |
\begin{example}
Suppose we start with the parent function \(f(x)=x^2\).
The figure below shows the graph of \(f(x)\) together with the shifted graphs
\(f(x)+1\) and \(f(x)-2\).
\[
\begin{array}{ccc}
\img{U1_10F11.jpg}{}{8em}{} & \img{U1_10F12.jpg}{}{8em}{} &
\img{U1_10F13.jpg}{}{8em}{}\\
x^2 & x^2+1 & x^2-2
\end{array}
\]
\end{example}
| Horizontal Shift |
|---|
| \(\begin{array}{c}
\mbox{Let \(h\) be a positive number. Then:}\\
\mbox{The graph of \(f(x+h)\) is the graph of \(f(x)\) shifted left by \(h\)} \\
\mbox{The graph of \(f(x-h)\) is the graph of \(f(x)\) shifted right by \(h\)}
\end{array}\) |
\begin{example}
We start with the parent function \(f(x)=|x|\). The figure below shows the
graph of \(f(x)\), \(f(x+1)\), and \(f(x-1)\).
\[
\begin{array}{ccc}
\img{U1_10F14.jpg}{}{8em}{} & \img{U1_10F15.jpg}{}{8em}{} &
\img{U1_10F16.jpg}{}{8em}{}\\
|x| & |x+1| & |x-1|
\end{array}
\]
\end{example}
\subsubsection{Reflections}
A reflected graph is obtained when either the input \(x\) or the output \(f(x)\)
is multiplied by \(-1\).
| Reflections |
|---|
| \(\begin{array}{c}
\mbox{The graph of \(-f(x)\) is the graph of \(f(x)\) reflected across the \(x\)-axis} \\
\mbox{The graph of \(f(-x)\) is the graph of \(f(x)\) reflected across the \(y\)-axis}
\end{array}\) |
\begin{example}
Let \(f(x)=\sqrt{x}\). The figure below shows the graph of \(f(x)\), \(-f(x)\) and
\(f(-x)\).
\[
\begin{array}{ccc}
\img{U1_10F17.jpg}{}{8em}{} & \img{U1_10F18.jpg}{}{8em}{}
& \img{U1_10F27.jpg}{}{8em}{}\\
\sqrt{x} & -\sqrt{x} & \sqrt{-x}
\end{array}
\]
\end{example}
\subsubsection{Shrinking and stretching}
Multiplication of the output \(f(x)\) by a positive number will
result in a stretched or compressed graph in the vertical direction.
So, unlike the shifting and reflection
(where the shape of the graph
remains unchanged), this time the graph will be distorted.
\begin{example}
See the figure for an example of a graph that is first stretched
vertically by \(2\), and then compressed vertically by \(2\).
\[
\begin{array}{ccc}
\img{U1_11F1.jpg}{}{10em}{} & \img{U1_11F2.jpg}{}{10em}{}
& \img{U1_11F3.jpg}{}{10em}{}\\
f(x) & 2f(x) & \displaystyle{\frac{1}{2} f(x)}
\end{array}
\]
\end{example}
| Vertical stretching or compressing (\(a> 1\) ) |
|---|
| \(\begin{array}{c}
\mbox{The graph of \(af(x)\) is the graph of \(f(x)\) stretched vertically by \(a\)} \\
\mbox{The graph of \(\dfrac{1}{a}f(x)\) is the graph of \(f(x)\) compressed vertically by \(a\)}
\end{array}\) |
\begin{example}
Suppose we want to draw the graph of the function \(f(x)=2x^2\).
The parent function is the square function \(x^2\). The graph is stretched vertically by 2.
This means that every \(y\) coordinate is multiplied by 2.
The figure below shows the graph of the parent function as a dashed curve, and the
graph of the stretched function as a solid curve.
\[\img{U1_11F4.jpg}{}{12em}{}\]
Note that if we fix an \(x\)-value,
the point on the stretched graph has a \(y\)-coordinate that is
twice the one on the original graph. For example, for \(x=1\) the black dot on the
original graph is \((1,1)\), and the red dot on the stretched graph is \((1,2)\).
\end{example}
Pay attention to the difference between stretching and shifting: when we shift a graph up by 2, we use addition,
so every output is increased by 2, including the zero output. But with stretching, we use multiplication, so the
zero output will remain the same, because zero times any number is still zero. So in the last example the
vertex of the stretched graph is still at \((0,0)\).
Multiplying the input \(x\) by a positive number stretches or compresses the graph
in the
horizontal direction. As before, the graph will be distorted.
\begin{example}
See the figure for an example of a graph that is first compressed
horizontally by \(2\), and then stretched horizontally by \(2\).
\[
\begin{array}{ccc}
\img{U1_11F1.jpg}{}{10em}{} & \img{U1_11F8.jpg}{}{10em}{}
& \img{U1_11F9.jpg}{}{20em}{}\\
f(x) & f(2x) & f\left(\dfrac{x}{2}\right)
\end{array}
\]
\end{example}
Note that this time the stretching/compressing is reversed: the graph is compressed when
the input \(x\) is multiplied by 2, and stretched when \(x\) is divided by \(2\).
| Horizontal stretching or compressing (\(a> 1\) ) |
|---|
| \(\begin{array}{c}
\mbox{The graph of \(f(ax)\) is the graph of \(f(x)\) compressed horizontally by \(a\)} \\
\mbox{The graph of \(f\left(\dfrac{x}{a}\right)\) the graph of \(f(x)\) is stretched horizontally by \(a\)}
\end{array}\) |
\begin{example}
We want to draw the graph of the function \(f(x)=\sqrt{2x}\).
The parent function is the square root function \(\sqrt{x}\). The graph is compressed horizontally
by 2. This means that for each \(y\) value, the \(x\)-coordinate of the compressed
graph is half the \(x\)-coordinate of the original graph.
The figure below shows the graph of the parent function as a black dashed curve, and the
graph of the compressed function as a red solid curve.
\[\img{U1_11F10.jpg}{}{12em}{}\]
Note that if we fix a \(y\)-value,
the point on the compressed graph has an \(x\)-coordinate that is
half the one on the original graph. For example, for \(y=2\) the black dot on the
original graph is \((4,2)\), and the red dot on the compressed graph is \((2,2)\).
\end{example}
\subsection{Combining simple transformations}
Shifts, reflections and compressing or stretching can be combined.
\begin{example}
\
\begin{itemize}
\item\(f(x+3)-5\) is \(f(x)\) shifted left by \(3\) and down by \(5\).
\item
\(-f(x-1)+2\) is \(f(x)\) reflected across the \(x\)-axis, then shifted right by \(1\) and up by \(2\).
\item
\(3f(-x)+1\) is \(f(x)\) reflected across the \(y\)-axis, then stretched vertically by 3,
then shifted up by 1.
\end{itemize}
\end{example}
Pay attention to the order in which the transformations are done.
Interchanging the order of shifts and
reflections gives different results. Compare the two cases:
\begin{enumerate}
\item ``\(f(x)\) reflected across the \(x\)-axis, then shifted up by \(2\)'' means:
\[
\begin{array}{cccl}
f(x) & \longrightarrow & -f(x) & \mbox{reflect across \(x\) axis} \\
-f(x) & \longrightarrow & -f(x)+2 & \mbox{shift up by 2}
\end{array}
\]
\item ``\(f(x)\) shifted up by 2, then reflected across the \(x\)-axis'' means:
\[
\begin{array}{cccl}
f(x) & \longrightarrow & f(x)+2 & \mbox{shift up by 2} \\
f(x)+2 & \longrightarrow & -(f(x)+2)=-f(x)-2 & \mbox{reflect across the \(x\) axis}.
\end{array}
\]
\end{enumerate}
So we get different results, \(-f(x)+2\) in one case and \(-f(x)-2\) in the other case.
Because the reflections and compressing or stretching involve multiplication and the shifting
involves addition, always do the reflections and compressing or stretching first, then do the shifting as the
last step.
\begin{example}
We are given \(f(x)=x^2\). Suppose we want to draw the graph of \(-f(x+2)+1=-(x+2)^2+1\).
We see that there is a reflection across the \(x\) axis, a left shift by 2, and a shift up
by 1.
\[
\begin{array}{cccc}
& \img{U1_11F14.jpg}{}{12em}{} & \rightarrow &\img{U1_11F15.jpg}{}{12em}{}\\
&\mbox{Start with \(x^2\)} & & \mbox{Reflect across \(x\) axis:} -x^2 \\
\rightarrow & \img{U1_11F16.jpg}{}{12em}{} & \rightarrow & \img{U1_11F17.jpg}{}{12em}{}\\
& \mbox{Shift left by 2:} -(x+2)^2 & & \mbox{Shift up by 1:} -(x+2)^2 +1
\end{array}
\]
\end{example}
Problems
\problem
Match each graph with one of the equations:
\[
\begin{array}{ccc}
\img{U1_10F1.jpg}{}{12em}{} & \img{U1_10F9.jpg}{}{12em}{} &
\img{U1_10F5.jpg}{}{12em}{}\\
1 & 2 & 3 \\[2ex]
\img{U1_10F3.jpg}{}{12em}{} & \img{U1_10F6.jpg}{}{12em}{} &
\img{U1_10F7.jpg}{}{12em}{}\\
4 & 5 & 6 \\[2ex]
\img{U1_10F4.jpg}{}{12em}{} & \img{U1_10F8.jpg}{}{12em}{} &
\img{U1_10F10.jpg}{}{12em}{}\\
7 & 8 & 9
\end{array}
\]
\[
\begin{array}{ccccc}
\mbox{a. } y=\sqrt{x} & \mbox{b. } y=3 & \mbox{c. } y=x &
\mbox{d. } y=x^2 & \mbox{e. } \displaystyle{y=\frac{1}{x}} \\
\mbox{f. } y=\sqrt[3]{x} & \mbox{g. } y=|x| & \mbox{h. } y=-3 & \mbox{i. } y=x^3
\end{array}
\]
\begin{sol}
a7 \ \ b2 \ \ c1 \ \ d4\ \ e8 \ \ f5 \ \ g6 \ \ h9 \ \ i3
\end{sol}
\mproblem
Match each graph with one of the equations:
\[
\begin{array}{ccc}
\img{U1_10F9.jpg}{}{12em}{} & \img{U1_10F5.jpg}{}{12em}{} &
\img{U1_10F3.jpg}{}{12em}{}\\
1 & 2 & 3 \\[2ex]
\img{U1_10F4.jpg}{}{12em}{} & \img{U1_10F1.jpg}{}{12em}{} &
\img{U1_10F10.jpg}{}{12em}{}\\
4 & 5 & 6 \\[2ex]
\img{U1_10F7.jpg}{}{12em}{} & \img{U1_10F8.jpg}{}{12em}{} &
\img{U1_10F6.jpg}{}{12em}{}\\
7 & 8 & 9
\end{array}
\]
\[
\begin{array}{ccccc}
\mbox{a. } y=\sqrt{x} & \mbox{b. } y=3 & \mbox{c. } y=x &
\mbox{d. } y=x^2 & \mbox{e. } \displaystyle{y=\frac{1}{x}} \\
\mbox{f. } y=\sqrt[3]{x} & \mbox{g. } y=|x| & \mbox{h. } y=-3 & \mbox{i. } y=x^3
\end{array}
\]
\problem
Sketch the graph of the functions:
\[
\begin{array}{cc}
\mbox{a. } \displaystyle{f(x)=|x|-2} & \hspace{1cm} \mbox{b. } \displaystyle{g(x)=(x+1)^2}
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item
\(f(x)=|x|-2\) is the parent function \(|x|\) function shifted down by 2:
\[
\begin{array}{ccc}
\img{U1_10F21.jpg}{}{8em}{} & \mbox{shift down by 2} \longrightarrow&
\img{U1_10F22.jpg}{}{8em}{}\\
|x| & & |x|-2
\end{array}
\]
\item
\(g(x)=(x+1)^2\) is the parent function \(x^2\) shifted left by 1:
\[
\begin{array}{ccc}
\img{U1_10F23.jpg}{}{8em}{} &\mbox{shift left by 1} \longrightarrow &
\img{U1_10F24.jpg}{}{8em}{}\\
x^2 & & (x+1)^2
\end{array}
\]
\end{enumerate}
\end{sol}
\mproblem
Sketch the graph of the functions:
\[
\begin{array}{cc}
\mbox{a. } f(x)=\sqrt{x-2} & \hspace{1cm} \mbox{b. } g(x)=x^3+2
\end{array}
\]
\problem
Sketch the graph of the functions:
\[
\begin{array}{cc}
\mbox{a. } f(x)=-x^2 & \hspace{1cm} \mbox{b. } \displaystyle{g(x)=\sqrt[3]{-x}}
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item\(f(x)=-x^2\) is the square function reflected across the \(x\) axis:
\[
\begin{array}{ccc}
\img{U1_10F23.jpg}{}{8em}{} &\mbox{reflect across \(x\)-axis} \longrightarrow &
\img{U1_10F20.jpg}{}{8em}{}\\
x^2 & & -x^2
\end{array}
\]
\item
\(g(x)=\sqrt[3]{-x}\) is the cube root root function reflected across the \(y\)-axis:
\[
\begin{array}{ccc}
\img{U1_10F25.jpg}{}{8em}{} &\mbox{reflect across \(y\)-axis} \longrightarrow &
\img{U1_10F26.jpg}{}{8em}{}\\
\sqrt[3]{x} & & \sqrt[3]{-x}
\end{array}
\]
Note that reflecting \(\sqrt[3]{x}\) across the \(x\)-axis or across the \(y\)-axis
gives the same result. That is because \(\sqrt[3]{-x}=-\sqrt[3]{x}\).
\end{enumerate}
\end{sol}
\mproblem
Sketch the graph of the functions:
\[
\begin{array}{cc}
\mbox{a. } f(x)=(-x)^3 & \hspace{1cm} \mbox{b. } -|x|
\end{array}
\]
\problem
Sketch the graph of the function \(\displaystyle{g(x)=\frac{1}{2}\sqrt{x}}\)
\begin{sol}
The parent function is the square root function \(\sqrt{x}\),
and the graph is compressed vertically by 2.
This means that every \(y\)-coordinate is divided by 2.
\[\img{U1_11F5.jpg}{}{20em}{}\]
Note that the black dot \((4,2)\) on the original graph corresponds to the a red
dot on the compressed graph with half the \(y\)-coordinate, \((4,1)\).
\end{sol}
\mproblem
Sketch the graph of the functions:
\[
\begin{array}{cc}
\mbox{a. } f(x)=2|x| & \hspace{1cm} \mbox{b. } \displaystyle{g(x)=\frac{1}{2} x^2}
\end{array}
\]
\problem
Draw the graph of the function \(\displaystyle{g(x)=\left(\frac{x}{2}\right)^2}\)
\begin{sol}
The parent function is the square function \(x^2\),
and the graph is stretched horizontally by 2.
This means that every \(x\)-coordinate is multiplied by \(2\).
\[\img{U1_11F11.jpg}{}{15em}{}\]
Note that the black dot \((1,1)\) on the original graph corresponds to the a red
dot on the stretched graph with twice the \(x\)-coordinate, \((2,1)\).
\end{sol}
\mproblem
Sketch the graph of the functions:
\[
\begin{array}{cc}
\mbox{a. } f(x)=\sqrt[3]{2x} & \hspace{1cm}
\mbox{b. } \displaystyle{g(x)=\left(\frac{x}{2}\right)^3}
\end{array}
\]
\problem
Sketch the graph of the following functions:
\[
\begin{array}{ccc}
\mbox{a. } f(x)=(x-2)^2+1 & \hspace{0.3cm} \mbox{b. } g(x)=|x+1|-2 &
\hspace{0.3cm} \mbox{c. } h(x)=-\sqrt{x-2}+1
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item
\(f(x)=(x-2)^2+1\) is the square function \(x^2\) shifted right by \(2\) and up by \(1\).
\[
\begin{array}{ccccc}
\hspace{-3ex} \mbox{Start with \(x^2\)} & \hspace{-3ex}\longrightarrow & \hspace{-3ex} \mbox{Shift right by 2} &\hspace{-3ex}
\longrightarrow
& \hspace{-3ex} \mbox{Shift up by 1}\\
\img{U1_11F18.jpg}{}{10em}{} &\hspace{-3ex} & \hspace{-3ex} \img{U1_11F19.jpg}{}{10em}{}
&\hspace{-3ex} & \hspace{-3ex} \img{U1_11F20.jpg}{}{10em}{}\\
x^2 & & (x-2)^2 & & (x-2)^2+1
\end{array}
\]
\item \(g(x)=|x+1|-2\) is the absolute value function \(|x|\) shifted left by \(1\) and
down by \(2\).
\[
\begin{array}{ccccc}
\mbox{Start with \(|x|\)} &\hspace{-1ex} \longrightarrow &\hspace{-1ex} \mbox{Shift left by 1} &\hspace{-1ex}
\longrightarrow
&\hspace{-1ex} \mbox{Shift down by 2}\\
\img{U1_11F21.jpg}{}{10em}{} &\hspace{-1ex} &\hspace{-1ex} \img{U1_11F22.jpg}{}{10em}{}
&\hspace{-1ex} &\hspace{-1ex} \img{U1_11F23.jpg}{}{10em}{}\\
|x| &\hspace{-1ex} &\hspace{-1ex} |x+1| &\hspace{-1ex} &\hspace{-1ex} |x+1|-2
\end{array}
\]
\item
\(h(x)=-\sqrt{x-2}+1\) is the square root function \(\sqrt{x}\) reflected across the \(x\)-axis,
then shifted right by \(2\) and up by \(1\).
\[
\begin{array}{cccc}
& \mbox{Start with \(\sqrt{x}\)} & \longrightarrow & \mbox{Reflect across the \(x\)-axis}\\
&\img{U1_11F24.jpg}{}{10em}{}& & \img{U1_11F25.jpg}{}{10em}{}\\
& \sqrt{x} & & -\sqrt{x} \\[2ex]
\longrightarrow & \mbox{Shift right by 2} & \longrightarrow & \mbox{Shift up by 1}\\
& \img{U1_11F26.jpg}{}{10em}{} & & \img{U1_11F27.jpg}{}{10em}{}\\
& -\sqrt{x-2} & & -\sqrt{x-2}+1
\end{array}
\]
\end{enumerate}
\end{sol}
\mproblem
Sketch the graph of the functions:
\[
\begin{array}{ccc}
\mbox{a. } f(x)=-|x-1|+2 & \hspace{0.3cm} \mbox{b. } g(x)=(x+1)^2+1 & \hspace{0.3cm}
\mbox{c. } h(x)=\sqrt{-x}+2
\end{array}
\]