
## SectionVertical and Horizontal Shifts

In this section, we explore how certain changes in the formula for a function affect its graph. In particular, we will compare the graph of $y = f (x)$ with the graphs of

\begin{equation*} y = f (x) + k,~~~ \text{ and }~~~ y =f (x + h) \end{equation*}

for different values of the constants $k$ and $h\text{.}$ Such variations are called transformations of the graph.

### SubsectionVertical Shifts

Figure230 shows the graphs of $f (x) = x^2 + 4\text{,}$ $g(x) = x^2 - 4\text{,}$ and the basic parabola, $y = x^2\text{.}$ By comparing tables of values, we can see exactly how the graphs of $f$ and $g$ are related to the basic parabola.

##### Vertical Shift
 $x$ $-2$ $-1$ $~0~$ $~~1~~$ $~2~$ $y=x^2$ $4$ $1$ $0$ $1$ $4$ $f(x)=x^2+4$ $8$ $5$ $4$ $5$ $8$
 $x$ $-2$ $-1$ $0$ $1$ $2$ $y=x^2$ $4$ $1$ $0$ $1$ $4$ $g(x)=x^2-4$ $0$ $-3$ $-4$ $-3$ $0$

Each $y$-value in the table for $f (x)$ is four units greater than the corresponding $y$-value for the basic parabola $y=x^2 \text{.}$ Consequently, each point on the graph of $f (x)$ is four units higher than the corresponding point on the basic parabola, as shown by the arrows. Similarly, each point on the graph of $g(x)$ is four units lower than the corresponding point on the basic parabola.

The graphs of $y = f (x)$ and $y = g(x)$ are said to be translations (or shifts) of the graph of $y = x^2\text{.}$ They are shifted to a different location in the plane but retain the same size and shape as the original graph. In general, we have the following principles.

###### Vertical Shifts

Let $k\gt 0\text{.}$ Compared with the graph of $y = f (x)\text{,}$

1. the graph of $y=f(x)+k$ is shifted upward $k$ units, and
2. the graph of $y=f(x)-k$ is shifted downward $k$ units.
###### Example231

Graph the following functions.

1. $g(x) = \abs{x} + 3$ and $y=\abs{x}$ and discuss how they are related.
2. $h(x) = \dfrac{1}{x}- 2$ and $y=\frac{1}{x}$ and discuss how they are related.
Solution
1. The table shows that the $y$-values for $g(x)$ are each three units greater than the corresponding $y$-values for the absolute value function. The graph of $g(x) = \abs{x} + 3$ is a translation of the basic graph of $y = \abs{x}\text{,}$ shifted upward three units, as shown in Figure232.

 $x$ $-2$ $-1$ $0$ $1$ $2$ $y=\abs{x}$ $2$ $1$ $0$ $1$ $2$ $g(x)=\abs{x}+3$ $5$ $4$ $3$ $4$ $5$
2. The table shows that the $y$-values for $h(x)$ are each two units smaller than the corresponding $y$-values for $y = \dfrac{1}{x}\text{.}$ The graph of $h(x) = \dfrac{1}{x} - 2$ is a translation of the basic graph of $y = \dfrac{1}{x}\text{,}$ shifted downward two units, as shown in Figure233.

 $x$ $-2$ $-1$ $\dfrac{1}{2}$ $1$ $2$ $y=\dfrac{1}{x}$ $\dfrac{-1}{2}$ $-1$ $2$ $1$ $\dfrac{1}{2}$ $h(x)=\dfrac{1}{x}-2$ $\dfrac{-5}{2}$ $-3$ $0$ $-1$ $\dfrac{-3}{2}$
###### Example234

The function $E = f (h)$ graphed in Figure235 gives the amount of electrical power, in megawatts, drawn by a community from its local power plant as a function of time during a 24-hour period in 2002. Sketch a graph of $y = f (h) + 300$ and interpret its meaning.

Solution

The graph of $y = f (h) + 300$ is a vertical translation of the graph of $f\text{,}$

as shown in Figure236. At each hour of the day, or for each value of $h\text{,}$ the $y$-coordinate is 300 greater than on the graph of $f\text{.}$ So at each hour, the community is drawing 300 megawatts more power than in 2002.

An evaporative cooler, or swamp cooler, is an energy-efficient type of air conditioner used in dry climates. A typical swamp cooler can reduce the temperature inside a house by 15 degrees. Figure238a shows the graph of $T = f (t)\text{,}$ the temperature inside Kate's house $t$ hours after she turns on the swamp cooler. Write a formula in terms of $f$ for the function $g$ shown in Figure238b and, give a possible explanation of its meaning.

### SubsectionHorizontal Shift

Now consider the graphs of $f (x) = (x + 2)^2$ and $g(x) = (x - 2)^2$ shown in Figure239. Compared with the graph of the basic function $y = x^2\text{,}$ the graph of $f (x) = (x + 2)^2$ is shifted two units to the left, as shown by the arrows. You can see why this happens by studying the function values in the table.

Locate a particular $y$-value for $y = x^2\text{,}$ say, $y = 4\text{.}$ You must move two units to the left in the table to find the same $y$-value for $f (x)\text{,}$ as shown by the arrow. In fact, each $y$-value for $f (x)$ occurs two units to the left when compared to the same $y$-value for $y = x^2\text{.}$

Similarly, the graph of $g(x) = (x - 2)^2$ is shifted two units to the right compared to the graph of $y = x^2\text{.}$ In the table for $g\text{,}$ each $y$-value for $g(x)$ occurs two units to the right of the same $y$-value for $y = x^2\text{.}$ In general, we have the following principle.

###### Horizontal Shifts

Let $h\gt 0\text{.}$ Compared with the graph of $y = f (x)\text{,}$

1. the graph of $y = f (x + h)~ ~$ is shifted $h$ units to the left, and
2. the graph of $y = f (x - h)~ ~$ is shifted $h$ units to the right.
###### Note242

At first, the direction of a horizontal translation may seem counterintuitive. Look again at the tables above to help you see how the shift occurs.

###### Example243

Graph the following functions.

1. $g(x) =\sqrt{x + 1}$ and $y=\sqrt{x}$ and discuss how they are related.
2. $h(x) = \dfrac{1}{(x - 3)^2}$ and $y=\dfrac{1}{x^2}$ and discuss how they are related.
Solution
1. Consider the table of values for the function.

 $x$ $-1$ $0$ $1$ $2$ $3$ $y=\sqrt{x}$ undefined $0$ $1$ $1.414$ $1.732$ $y=\sqrt{x+1}$ $0$ $1$ $1.414$ $1.732$ $2$

The table shows that each $y$-value for $g(x)$ occurs one unit to the left of the same $y$-value for the graph of $y=\sqrt{x}\text{.}$ Consequently, each point on the graph of $y = g(x)$ is shifted one unit to the left of $y =\sqrt{x}\text{,}$ as shown in Figure244.

2. Consider the table of values for the function.

 $x$ $-1$ $0$ $1$ $2$ $3$ $4$ $y=\dfrac{1}{x}$ $1$ undefined $1$ $\dfrac{1}{4}$ $\dfrac{1}{9}$ $\dfrac{1}{16}$ $y=\dfrac{1}{(x-3)^2}$ $\dfrac{1}{16}$ $\dfrac{1}{9}$ $\dfrac{1}{4}$ $1$ undefined $1$

The table shows that each $y$-value for $h(x)$ occurs three units to the right of the same $y$-value for the graph of $y =\dfrac{1}{x^2}\text{.}$ Consequently, each point on the graph of $y = h(x)$ is shifted three units to the right of $y =\dfrac{1}{x^2}\text{,}$ as shown in Figure245.

###### Example246

The function $N = f (p)$ graphed in Figure247 gives the number of people who have a given eye pressure level $p$ measured in millimeters of mercury (mm Hg) from a sample of 100 people with healthy eyes, and the function $g$ gives the number of people with pressure level $p$ in a sample of 100 glaucoma patients.

1. Write a formula for $g$ as a transformation of $f\text{.}$
2. For what pressure readings could a doctor be fairly certain that a patient has glaucoma?
Solution
1. The graph of $g$ is translated $10$ units to the right of $f\text{,}$ so $g(p) = f (p - 10)\text{.}$
2. Pressure readings above $40$mm Hg are a strong indication of glaucoma. Readings between $10$ and $40$mm Hg cannot conclusively distinguish healthy eyes from those with glaucoma.

The function $C = f (t)$ in Figure249 gives the caffeine level in Delbert's bloodstream at time $t$ hours after he drinks a cup of coffee, and $g(t)$ gives the caffeine level in Francine's bloodstream. Write a formula for $g$ in terms of $f\text{,}$ and explain what it tells you about Delbert and Francine.

###### Example250

Graph $f (x) = (x + 4)^3 + 2\text{.}$

Solution

We identify the basic graph from the structure of the formula for $f (x)\text{.}$ In this case, the basic graph is $y = x^3\text{,}$ so we begin by locating a few points on that graph, as shown in Figure251.

We will perform the translations separately, following the order of operations. First, we sketch a graph of $y = (x + 4)^3$ by shifting each point on the basic graph four units to the left. We then move each point up two units to obtain the graph of $f (x) = (x + 4)^3 + 2\text{.}$ All three graphs are shown in Figure251.

1. Graph the function $f (x) = \abs{x - 2} - 1\text{.}$
2. How is the graph of $f$ different from the graph of $y=\abs{x}\text{?}$