Exercise 1.3.9 (9-26)

Question

Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions given in Table 1.2.3, and then applying the appropriate transformations.

Khagan Gulusoy · Accepted Answer

\begin{enumerate}
    \item [9.] We can draw the graph of $y=1+x^2$ by shifting the graph of $y=x^2$ upward by 1 unit.\begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-4:4, samples=50, mark=none, color=red, dashed, ultra thick] {x^2};
         \addplot[domain=-4:4, samples=150, mark=none, color=violet, ultra thick] {x^2+1};
         \legend{$x^2$, $1+x^2$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [10.] We draw the graph of  $y=(x+1)^2$ by moving the graph of $y=x^2$ by $1$ unit to the left. \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-4:3, samples=50, mark=none, color=red, dashed, ultra thick] {x^2};
         \addplot[domain=-4:3, samples=50, mark=none, color=violet, ultra thick] {(x+1)^2};
         \legend{$x^2$, $(x+1)^2$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [11.] We draw the graph of  $y=|x+2|$ by moving the graph of $y=|x|$ by $2$ unit to the left. \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-4:4, samples=50, mark=none, color=red, dashed, ultra thick] {abs(x)};
         \addplot[domain=-4:3, samples=50, mark=none, color=violet, ultra thick] {abs(x+2)};
         \legend{$|x|$, $|x+2|$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}
    \item [12.] We draw the graph of $y=1-x^3$ as follows:
    \begin{itemize}
        \item We reflect the graph of $y=x^3$ about the $x$-axis and get the graph of $y=-x^3;$
        \item We shift the graph of $y=-x^3$ by $1$ unit and get the graph of $y=1-x^3.$
    \end{itemize}
    Thus, we sketch the graph of $y=1-x^3$ as follows:
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-2:2, samples=50, mark=none, color=red, dashed, ultra thick] {x^3};
         \addplot[domain=-2:2, samples=50, mark=none, color=blue, dashed, ultra thick] {-x^3};
         \addplot[domain=-2:2, samples=50, mark=none, color=violet, ultra thick] {-x^3+1};
         \legend{$x^3$, $-x^3$, $-x^3+1$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}
    
    \item [13.] We can draw the graph of $y=\dfrac{1}{x}+2$ by shifting the graph of $y=\dfrac{1}{x}$ upward by 2 units. 
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, ymax=20, ymin=-20, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-1:-0.001, samples=50, mark=none, color=red, dashed, ultra thick] {1/x};
         \addlegendentry{$1/x$};
         \addplot[domain=-1:-0.001, samples=50, mark=none, color=violet, ultra thick] {1/x+2};
         \addlegendentry{$1/x+2$};
         \addplot[domain=0.001:1, samples=50, mark=none, color=red, dashed, ultra thick] {1/x};
         \addplot[domain=0.001:1, samples=50, mark=none, color=violet, ultra thick] {1/x+2};
         \addplot [dashed, domain=-1:1, lightgray, thin] coordinates {(-1,2) (1,2)};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [14.] Notice that $-\sqrt{x}-1=-(\sqrt{x}+1).$ Thus, to draw the graph of $y=-\sqrt{x}-1$, we use the following logic:
    \begin{itemize}
        \item First of all, we must shift the graph of $y=\sqrt{x}$ by $1$ unit upward to get the graph of $y=\sqrt{x}+1;$
        \item After that, we reflect the graph of $y=\sqrt{x}+1$ about the $x$-axis to get the graph of $y=-\sqrt{x}-1.$
    \end{itemize}
    Thus, we sketch the graph of $y=1-x^3$ as follows:
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=0:16, samples=50, mark=none, color=red, dashed, ultra thick] {sqrt(x)};
         \addplot[domain=0:16, samples=50, mark=none, color=blue, dashed, ultra thick] {sqrt(x)+1};
         \addplot[domain=0:16, samples=50, mark=none, color=violet, ultra thick] {-1*sqrt(x)-1};
         \legend{$\sqrt{x}$, $\sqrt{x}+1$, $-\sqrt{x}-1$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [15.] To draw the graph $y=\sin (4x)$ we need to shrink the graph of $y=\sin x$ by a factor of $4$ as follows:
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-4:4, samples=50, mark=none, color=red, dashed, ultra thick] {sin(deg(x))};
         \addplot[domain=-4:4, samples=150, mark=none, color=violet, ultra thick] {sin(deg(4*x))};
         \legend{$\sin x$, $\sin 4x$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [16.] We can draw the graph of $y=1+\dfrac{1}{x^2}$ by shifting the graph of $y=1+\dfrac{1}{x^2}$ upward by 1 unit. 
    \begin{figure}[ht]\centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, ymax=7, ymin=0, x label style={at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=0.1:2, samples=200, mark=none, color=red, dashed, ultra thick] {1/x^2};
         \addlegendentry{$1/x^2$};
         \addplot[domain=0.1:2, samples=200, mark=none, color=violet, ultra thick] {1+1/x^2};
         \addlegendentry{$1+1/x^2$};
         \addplot[domain=-2:-0.1, samples=200, mark=none, color=red, dashed, ultra thick] {1/x^2};
         \addplot[domain=-2:-0.1, samples=200, mark=none, color=violet, ultra thick] {1+1/x^2};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [17.] To draw the graph of $2+\sqrt{x+1},$ we follow the following logic:
    \begin{itemize}
        \item We move the graph of $y=\sqrt{x}$ by $1$ unit to the left.
        \item Then, we shift the graph of $\sqrt{x+1}$ by $2$ units upward.
    \end{itemize}
    Thus, we sketch the graph of $y=2+\sqrt{x+1}$ as follows:
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, xmin=-2, xtick={-1}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=0:16, samples=50, mark=none, color=red, dashed, ultra thick] {sqrt(x)};
         \addplot[domain=-1:16, samples=50, mark=none, color=blue, dashed, ultra thick] {sqrt(x+1)};
         \addplot[domain=-1:16, samples=50, mark=none, color=violet, ultra thick] {sqrt(x+1)+2};
         \legend{$\sqrt{x}$, $\sqrt{x+1}$, $\sqrt{x+1}+2$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [18.] To draw the graph of $y=-(x-1)^2+3,$ we follow the following logic:
    \begin{itemize}
        \item We move the graph of $y=x^2$ by $1$ unit to the right. We get the graph of $y=(x-1)^2.$
        \item Then, we reflect the graph of $y=(x-1)^2$ about the $x$-axis to get the graph of $y=-(x-1)^2$.
        \item To get the graph of $y=-(x-1)^2+3,$ we must shift the graph of $y=-(x-1)^2$ by 3 units upward.
    \end{itemize}
    Thus, we sketch the graph of $y=-(x-1)^2+3$ as follows:
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-4:4, samples=50, mark=none, color=red, dashed, ultra thick] {x^2};
         \addplot[domain=-4:4, samples=50, mark=none, color=blue, dashed, ultra thick] {(x-1)^2};
         \addplot[domain=-4:4, samples=50, mark=none, color=green!60!black, dashed, ultra thick] {-(x-1)^2};
         \addplot[domain=-4:4, samples=50, mark=none, color=violet, ultra thick] {-(x-1)^2+3};
         \legend{$x^2$, $(x-1)^2$, $-(x-1)^2$, $-(x-1)^2+3$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [19.] Notice that $x^2-2x+5=x^2-2x+1+4=(x-1)^2+4.$ Thus, to draw the graph of $y=x^2-2x+5$, we use the following logic:
    \begin{itemize}
        \item We move the graph of $y=x^2$ by $1$ unit to the right. We get the graph of $y=(x-1)^2.$
        \item To get the graph of $y=(x-1)^2+4,$ we must shift the graph of $y=(x-1)^2$ by 4 units upward.
    \end{itemize}
    Thus, we sketch the graph of $y=(x-1)^2+4$ as follows:
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-4:4, samples=50, mark=none, color=red, dashed, ultra thick] {x^2};
         \addplot[domain=-4:4, samples=50, mark=none, color=blue, dashed, ultra thick] {(x-1)^2};
         \addplot[domain=-4:4, samples=50, mark=none, color=violet, ultra thick] {(x-1)^2+4};
         \legend{$x^2$, $(x-1)^2$, $(x-1)^2+4$};
         \addplot [dashed, domain=-1:1, gray, ultra thick] coordinates {(1,0) (1,4)};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [20.] To draw the graph of $y=(x+1)^3+2$, we use the following logic:
    \begin{itemize}
        \item We move the graph of $y=x^3$ by $1$ unit to the left. We get the graph of $y=(x+1)^3.$
        \item To get the graph of $y=(x+1)^3+2,$ we must shift the graph of $y=(x+1)^3$ by 2 units upward.
    \end{itemize}
    Thus, we sketch the graph $y=(x+1)^3+2$ as follows:
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-3:3, samples=50, mark=none, color=red, dashed, ultra thick] {x^3};
         \addplot[domain=-3:3, samples=50, mark=none, color=blue, dashed, ultra thick] {(x+1)^3};
         \addplot[domain=-3:3, samples=50, mark=none, color=violet, ultra thick] {(x+1)^3+2};
         \legend{$x^3$, $(x+1)^3$, $(x+1)^3+2$};
         \addplot [dashed, domain=-1:1, gray, ultra thick] coordinates {(-1,0) (-1,2)};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [21.] To draw the graph of $y=2-|x|$, we use the following logic:
    \begin{itemize}
        \item We reflect the graph of $y=|x|$ about the $x$-axis. We get the graph of $y=-|x|.$
        \item To get the graph of $y=2-|x|,$ we must shift the graph of $y=-|x|$ by 2 units upward.
    \end{itemize}
    Thus, we sketch the graph $y=2-|x|$ as follows:
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-3:3, samples=50, mark=none, color=red, dashed, ultra thick] {abs(x)};
         \addplot[domain=-3:3, samples=50, mark=none, color=blue, dashed, ultra thick] {-1*abs(x)};
         \addplot[domain=-3:3, samples=50, mark=none, color=violet, ultra thick] {2-abs(x)};
         \legend{$|x|$, $-|x|$, $2-|x|$};
         %\addplot [dashed, domain=-1:1, gray, ultra thick] coordinates {(-1,0) (-1,2)};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [22.] To draw the graph of $y=2-2\cos x,$ we use the following logic:
    \begin{itemize}
        \item First of all, we reflect the graph of the cosine function about the $x$-axis and get the graph of $y=-\cos x.$
        \item Then, we stretch the graph vertically by a factor of $2$ and get the graph of $y=-2\cos x.$
        \item Finally, we shift the graph of $y=-2\cos x$ by 2 units upward, getting the desired graph of $y=2-2\cos x.$
    \end{itemize}
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-4:4, samples=50, mark=none, color=red, dashed, ultra thick] {cos(deg(x))};
         \addplot[domain=-4:4, samples=50, mark=none, color=blue, dashed, ultra thick] {-cos(deg(x))};
         \addplot[domain=-4:4, samples=50, mark=none, color=green!60!black, dashed, ultra thick] {-2*cos(deg(x))};
         \addplot[domain=-4:4, samples=50, mark=none, color=violet, ultra thick] {2-2*cos(deg(x))};
         \legend{$\cos x$, $-\cos x$, $-2\cos x$, $2-2\cos x$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [23.] To draw the graph of $y=3\sin \dfrac{1}{2} x+1,$ we use the following logic:
    \begin{itemize}
        \item First of all, we stretch the graph of the sine function horizontally by a factor of $2.$
        \item Then, we stretch the graph vertically by a factor of $3$ and get the graph of $y=3\sin \dfrac{1}{2}x.$
        \item Finally, we shift the graph of $y=3\sin \dfrac{1}{2}x$ by 1 units upward, getting the desired graph of $y=3\sin \dfrac{1}{2}x+1.$
    \end{itemize}
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}, legend style={at={(0.2,0.95)}, anchor=north}]
         \addplot[domain=-5:5, samples=50, mark=none, color=red, dashed, ultra thick] {sin(deg(x))};
         \addplot[domain=-5:5, samples=50, mark=none, color=blue, dashed, ultra thick] {sin(deg(0.5*x))};
         \addplot[domain=-5:5, samples=50, mark=none, color=green!60!black, dashed, ultra thick] {3*sin(deg(0.5*x))};
         \addplot[domain=-5:5, samples=50, mark=none, color=violet, ultra thick] {3*sin(deg(0.5*x))+1};
         \legend{$\sin x$, $\sin 1/2\ x$, $3\sin 1/2 \ x$, $3\sin 1/2 \ x+1$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [24.] To draw the graph of $y=\dfrac{1}{4}	an \left(x-\dfrac{\pi}{4}ight),$ we use the following logic:
    \begin{itemize}
        \item First of all, we move the graph of $y=	an x$ by $\dfrac{\pi}{4}$ units to the right and get the graph of $y=	an \left(x-\dfrac{\pi}{4}ight).$ 
        \item Then, we shrink the graph vertically by a factor of $4$ and get the graph of $y=\dfrac{1}{4}	an \left(x-\dfrac{\pi}{4}ight).$
    \end{itemize}
    Thus, we sketch the graph of $y=\dfrac{1}{4}	an \left(x-\dfrac{\pi}{4}ight)$ as follows:

\begin{figure}[ht] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, domain=-pi/2 + 0.01:pi/2 - 0.01, samples=200, legend style={at={(1.1,0.35)}, anchor=north east,},]
         \addplot[mark=none, color=red, ultra thick, restrict y to domain=-5:5, dashed] {tan(deg(x))};
         \addlegendentry{$	an(x)$};
         \addplot[mark=none, color=blue, ultra thick, restrict y to domain=-5:5, dashed] {tan(deg(x - pi/4))};
         \addlegendentry{$	an(x - \frac{\pi}{4})$};
         \addplot[mark=none, color=violet, ultra thick, restrict y to domain=-5:5] {0.25*tan(deg(x - pi/4))};
         \addlegendentry{$0.25 	an(x - \frac{\pi}{4})$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [25.] To draw the graph of $y=|\cos x\pi|,$ we use the following logic:
    \begin{itemize}
        \item First of all, we shrink the graph of the cosine function horizontally by a factor of $\pi$ to obtain the graph of $y=\cos \pi x.$
        \item Since the value of $\cos x\pi$ is negative when $x\in \left[2\pi n+\dfrac{1}{2}\pi, \ 2\pi n + \dfrac{3}{2}\piight]$ for some $n\in \mathbb{Z},$ we reflect that part of the graph about the $x$-axis to obtain the graph of $y=|\cos x\pi|.$
    \end{itemize}
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm},width=6.5cm]
         \addplot[domain=-4:4, samples=50, mark=none, color=red, dashed, ultra thick] {cos(deg(x))};
         \legend{$\cos x$};
    \end{axis}    
    \end{tikzpicture}
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}, width=6.5cm]
         \addplot[domain=-4:4, samples=200, mark=none, color=blue, dashed, ultra thick] {cos(deg(pi*x))};
         \legend{$\cos \pi x$};
    \end{axis}    
    \end{tikzpicture}
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm},width=6.5cm]
         \addplot[domain=-4:4, samples=1000, mark=none, color=violet, ultra thick] {abs(cos(deg(pi*x)))};
         \legend{$|\cos \pi x|$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item [26.] To draw the graph of $y=|\sqrt{x}-1|,$ we use the following logic:
    \begin{itemize}
        \item First of all, we shift the graph of $y=\sqrt{x}$ by $1$ unit downward to obtain the graph of $y=\sqrt{x}-1.$
        \item Since the function gets negative when $x\in [0,1),$ we reflect this part about the $x$-axis. This way, we get the desired graph of $y=|\sqrt{x}-1|.$
    \end{itemize}
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm},width=4.5cm]
         \addplot[domain=0:5, samples=50, mark=none, color=red, dashed, ultra thick] {sqrt(x)};
         \legend{$\sqrt{x}$};
    \end{axis}    
    \end{tikzpicture}
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}, width=4.5cm]
         \addplot[domain=0:5, samples=50, mark=none, color=blue, dashed, ultra thick] {sqrt(x)-1};
         \legend{$\sqrt{x}-1$};
    \end{axis}    
    \end{tikzpicture}
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm},width=4.5cm]
         \addplot[domain=0:5, samples=1000, mark=none, color=violet, ultra thick] {abs(sqrt(x)-1)};
         \legend{$|\sqrt{x}-1|$};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}
    
\end{enumerate}

Exercise 1.3.9 (9-26)

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Exercise 1.3.9 (9-26)

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