5.2: Transformation of Graphs
 Page ID
 48976
For a given function, we now study how the graph of the function changes when performing elementary operations, such as adding, subtracting, or multiplying a constant number to the input or output. We will study the behavior in five specific examples.
 Consider the following graphs:
We see that the function \(y=x^2\) is shifted up by \(2\) units, respectively down by \(2\) units. In general, we have:
Consider the graph of a function \(y=f(x)\). Then, the graph of \(y=f(x)+c\) is that of \(y=f(x)\) shifted up or down by \(c\). If \(c\) is positive, the graph is shifted up, if \(c\) is negative, the graph is shifted down.
 Next, we consider the transformation of \(y=x^2\) given by adding or subtracting a constant to the input \(x\).
Now, we see that the function is shifted to the left or right. Note, that \(y=(x+1)^2\) shifts the function to the left, which can be seen to be correct, since the input \(x=1\) gives the output \(y=((1)+1)^2=0^2=0\).
Consider the graph of a function \(y=f(x)\). Then, the graph of \(y=f(x+c)\) is that of \(y=f(x)\) shifted to the left or right by \(c\). If \(c\) is positive, the graph is shifted to the left, if \(c\) is negative, the graph is shifted to the right.
 Another transformation is given by multiplying the function by a fixed positive factor.
This time, the function is stretched away or compressed towards the \(x\)axis.
Consider the graph of a function \(y=f(x)\) and let \(c>0\). Then, the graph of \(y=c\cdot f(x)\) is that of \(y=f(x)\) stretched away or compressed towards the \(x\)axis by a factor \(c\). If \(c>1\), the graph is stretched away from the \(x\)axis, if \(0<c<1\), the graph is compressed towards the \(x\)axis.
 Similarly, we can multiply the input by a positive factor.
This time, the function is stretched away or compressed towards the \(y\)axis.
Consider the graph of a function \(y=f(x)\) and let \(c>0\). Then, the graph of \(y=f(c\cdot x)\) is that of \(y=f(x)\) stretched away or compressed towards the \(y\)axis by a factor \(c\). If \(c>1\), the graph is compressed towards the \(y\)axis, if \(0<c<1\), the graph is stretched away from the \(y\)axis.
 The last transformation is given by multiplying \((1)\) to the input or output, as displayed in the following chart.
Here, the function is reflected either about the \(x\)axis or the \(y\)axis.
Consider the graph of a function \(y=f(x)\). Then, the graph of \(y=f(x)\) is that of \(y=f(x)\) reflected about the \(x\)axis. Furthermore, the graph of \(y=f(x)\) is that of \(y=f(x)\) reflected about the \(y\)axis.
Guess the formula for the function, based on the basic graphs in Section 5.1 and the transformations described above.
Solution
 This is the squareroot function shifted to the left by \(2\). Thus, by Observation, this is the function \(f(x)=\sqrt{x+2}\).
 This is the graph of \(y=\dfrac 1 x\) reflected about the \(x\)axis (or also \(y=\dfrac 1 x\) reflected about the \(y\)axis). In either case, we obtain the rule \(y=\dfrac 1 x\).

This is a parabola reflected about the \(x\)axis and then shifted up by \(3\). Thus, we get: \[\begin{aligned}
&y=x^{2} \\
\text{reflecting about the xaxis gives }&y=x^{2} \\
\text{shifting the graph up by 3 gives }&y=x^{2}+3
\end{aligned} \nonumber \] 
Starting from the graph of the cubic equation \(y=x^3\), we need to reflect about the \(x\)axis (or also \(y\)axis), then shift up by \(2\) and to the right by \(3\). These transformations affect the formula as follows: \[\begin{aligned}
& y=x^{3} \\
\text {reflecting about the xaxis gives } & y=x^{3}\\
\text{shifting up by 2 gives } & y=x^{3}+2\\
\text{shifting the the right by 3 gives } & y=(x3)^{3}+2
\end{aligned} \nonumber \]
All these answers can be checked by graphing the function with the TI84.
Sketch the graph of the function, based on the basic graphs in Section 5.1 and the transformations described above.
 \(y=x^2+3\)
 \(y=(x+2)^2\)
 \(y=x32\)
 \(y=2\cdot \sqrt{x+1}\)
 \(y=\left(\dfrac 1 {x}+2\right)\)
 \(y=(x+1)^3\)
Solution
 This is the parabola \(y=x^2\) shifted up by \(3\). The graph is shown below.
 \(y=(x+2)^2\) is the parabola \(y=x^2\) shifted \(2\) units to the left.
 The graph of the function \(f(x)=x32\) is the absolute value shifted to the right by \(3\) and down by \(2\). (Alternatively, we can first shift down by \(2\) and then to the right by \(3\).)
 Similarly, to get from the graph of \(y=\sqrt{x}\) to the graph of \(y=\sqrt{x+1}\), we shift the graph to the left, and then for \(y=2\cdot \sqrt{x+1}\), we need to stretch the graph by a factor \(2\) away from the \(x\)axis. (Alternatively, we could first stretch the the graph away from the \(x\)axis, then shift the graph by \(1\) to the left.)
 For \(y=\left(\dfrac 1 x +2\right)\), we start with \(y=\dfrac 1 x\) and add \(2\), giving \(y=\dfrac 1 x +2\), which shifts the graph up by \(2\). Then, taking the negative gives \(y=\left(\dfrac 1 x +2\right)\), which corresponds to reflecting the graph about the \(x\)axis. Note, that in this case, we cannot perform these transformations in the opposite order, since the negative of \(y=\dfrac 1 x\) gives \(y=\dfrac 1 x\), and adding \(2\) gives \(y=\dfrac 1 x+2\) which is not equal to \((\dfrac 1 x+2)\).
 We start with \(y=x^3\). Adding \(1\) in the argument, \(y=(x+1)^3\), shifts its graph to the left by \(1\). Then, taking the negative in the argument gives \(y=(x+1)^3\), which reflects the graph about the \(y\)axis. Here, the order in which we perform these transformations is again important. In fact, if we first take the negative in the argument, we obtain \(y=(x)^3\). Then, adding one in the argument would give \(y=((x+1))^3=(x1)^3\) which is different than our given function \(y=(x+1)^3\).
All these solutions may also easily be checked by using the graphing function of the calculator.
 The graph of \(f(x)=x^35\) is stretched away from the \(y\)axis by a factor of \(3\). What is the formula for the new function?
 The graph of \(f(x)=\sqrt{6x^2+3}\) is shifted up \(5\) units, and then reflected about the \(x\)axis. What is the formula for the new function?
 How are the graphs of \(y=2x^3+5x9\) and \(y=2(x2)^3+5(x2)9\) related?
 How are the graphs of \(y=(x2)^2\) and \(y=(x+3)^2\) related?
Solution
 By Observation on page , we have to multiply the argument by \(\frac 1 3\). The new function is therefore:
\[f\Big(\dfrac 1 3 \cdot x\Big)=\left\Big(\dfrac 1 3\cdot x \Big)^35\right=\left\dfrac 1 {27} \cdot x^35\right \nonumber\]
 After the shift, we have the graph of a new function \(y=\sqrt{6x^2+3}+5\).Then, a reflection about the \(x\)axis gives the graph of the function \(y=(\sqrt{6x^2+3}+5)\).
 By Observation on page , we see that we need to shift the graph of \(y=2x^3+5x9\) by \(2\) units to the right.

The formulas can be transformed into each other as follows: \[\begin{align*}
\text{We begin with } & y=(x2)^{2}\\
\text{Replacing x by x+5 gives } & y=((x+5)2)^{2}=(x+3)^{2}\\
\text{Replacing x by x gives } & y=((x)+3)^{2}=(x+3)^{2}
\end{align*} \nonumber \]Therefore, we have performed a shift to the left by \(5\), and then a reflection about the \(y\)axis.We want to point out that there is a second solution for this problem: We begin with & \(y=(x2)^2\).
Replacing \(x\) by \(x\) gives \(y=((x)2)^2=(x2)^2\).
Replacing \(x\) by \(x5\) gives \(y=((x5)2)^2=(x+52)^2=(x+3)^2\). Therefore, we could also first perform a reflection about the \(y\)axis, and then shift the graph to the right by \(5\).
Some of the above functions have special symmetries, which we investigate now.
A function \(f\) is called even if \(f(x)=f(x)\) for all \(x\).
Similarly, a function \(f\) is called odd if \(f(x)=f(x)\) for all \(x\).
Determine, if the following functions are even, odd, or neither. \(f(x)=x^2\), \(g(x)=x^3\), & \(h(x)=x^4\), \(k(x)=x^5\),
\(l(x)=4x^5+7x^32x\), & \(m(x)=x^2+5x\).
Solution
The function \(f(x)=x^2\) is even, since \(f(x)=(x)^2=x^2\). Similarly, \(g(x)=x^3\) is odd, \(h(x)=x^4\) is even, and \(k(x)=x^5\) is odd, since
\[\begin{aligned} g(x)&=(x)^3=x^3=g(x) \\ h(x)&=(x)^4=x^4=h(x) \\ k(x)&=(x)^5=x^5=k(x) \end{aligned}\]
Indeed, we see that a function \(y=x^n\) is even, precisely when \(n\) is even, and \(y=x^n\) is odd, precisely when \(n\) is odd. (These examples are in fact the motivation behind defining even and odd functions as in Definition EvenOdd above.)
Next, in order to determine if the function \(l\) is even or odd, we calculate \(l(x)\) and compare it with \(l(x)\).
\[\begin{aligned} l(x)&=4(x)^5+7(x)^32(x)\\ &=4x^57x^3+2x \\ &=(4x^5+7x^32x)\\ &=l(x) \end{aligned}\]
Therefore, \(l\) is an odd function.
Finally, for \(m(x)=x^2+5x\), we calculate \(m(x)\) as follows:
\[m(x)=(x)^2+5(x)=x^25x \nonumber \]
Note, that \(m\) is not an even function, since \(x^25x\neq x^2+5x\). Furthermore, \(m\) is also not an odd function, since \(x^25x\neq (x^2+5x)\). Therefore, \(m\) is a function that is neither even nor odd.
An even function \(f\) is symmetric with respect to the \(y\)axis (if you reflect the graph of \(f\) about the \(y\)axis you get the same graph back), since even functions satisfy \(f(x)=f(x)\):
Example \(y=x^2\):
An odd function \(f\) is symmetric with respect to the origin (if you reflect the graph of \(f\) about the \(y\)axis and then about the \(x\)axis you get the same graph back), since odd functions satisfy \(f(x)=f(x)\):
Example \(y=x^3\):