Understand and create graphs of functions.
Identify \(x-\) and \(y-\) intercepts.
Understand and create vertical and horizontal translations of functions.
Use inequality and interval notation.
Consider the following graph:
If you were told to “Draw the graph of \(y = x^2\)” you would probably remember how to plot points and draw the shape.
But suppose I asked you this instead: “Here's a function, \(y = x^2\text{.}\) And here's a shape, that sort of looks like a U. What do they actually have to do with each other?” This is a harder question! What does it mean to graph a function?
The answer is simple, but it has important implications for a proper understanding of functions. Recall that every point on the plane is designated by a unique \((x, y)\) pair of coordinates: for instance, one point is \((5, 3)\text{.}\) We say that its \(x\)-value is \(5\) and its \(y\)-value is \(3\text{.}\)
A few of these points have the particular property that their \(y\)-values are the square of their \(x\)-values. For instance, the points \((0, 0)\text{,}\) \((3, 9)\text{,}\) and \((-5, 25)\) all have that property. \((5, 3)\) and \((-2,-4)\) do not.
The graph shown-the pseudo-U shape-is all the points in the plane that have this property. Any point whose \(y\)-value is the square of its \(x\)-value is on this shape; any point whose \(y\)-value is not the square of its \(x\)-value is not on this shape. Hence, glancing at this shape gives us a complete visual picture of the function \(y = x^2\) if we know how to interpret it correctly.
Remember that every function specifies a relationship between two variables. When we graph a function, we put the independent variable on the horizontal axis, and the dependent variable on the vertical axis.
For instance, recall the function that describes Alice's money as a function of her hours worked. Since Alice makes \(\$12\) per hour, her financial function is \(m(t) = 12t\text{.}\) We can graph it like this.
This simple graph has a great deal to tell us about Alice's job, if we read it correctly.
The graph contains the point \((2, 24).\) What does that tell us? That after Alice has worked for two hours, she has made \(\$24.\)
The graph goes through the origin (the point \((0, 0).\)) What does that tell us? That when she works 0 hours, Alice makes no money.
The graph exists only in the first quadrant. What does that tell us? On the mathematical level, it indicates the domain of the function (\(t \geq 0\)) and the range of the function (\(m \geq 0\)). In terms of the situation, it tells us that Alice cannot work negative hours or make negative money.
The graph is a straight line. What does that tell us? That Alice makes the same amount of money every hour: every hour she works, her money goes up by \(\$12.\) (\(\$12\) per hour is the slope of the line-more on this in the section on linear functions.)
Consider now the following, more complicated graph, which represents Alice's hair length as a function of time (where time is now measured in weeks instead of hours).
What does this graph \(h(t)\) tell us? We can start with the same sort of simple analysis.
The graph goes through the point \((0, 12)\text{.}\)This tells us that at time (\(t = 0\)), Alice's hair is \(12''\) long.
The range of this graph appears to be \(12 \leq h \leq 18\text{.}\) Alice never allows her hair to be shorter than \(12''\) or longer than \(18''\text{.}\)
But what about the shape of the graph? The graph shows a gradual incline up to \(18\text{,}\) and then a precipitous drop back down to \(12\text{;}\) and this pattern repeats throughout the shown time. The most likely explanation is that Alice's hair grows slowly until it reaches \(18''\text{,}\) at which point she goes to the hair stylist and has it cut, within a very short time (an hour or so), to \(12''\text{.}\) Then the gradual growth begins again.
Certain points on the graph of a function have special characteristics. Intercepts are points on the graph which are on an axis. In the graph of \(y = f(x),\) the \(x\)-intercepts are \((-4,0),\) \((1,0)\) and \((3.5,0).\) There is only one \(y\)-intercept at \((0,-2).\)
Functions don't always have both kinds of intercepts. The graph of \(y = h(x)\) shown below has only a \(y\)-intercept at the point \((0,1)\) and no \(x\)-intercepts.
Sometimes the intercepts are at the same point. In the graph shown below, the point \((0,0)\) is both a \(y\)-intercept and an \(x\)-intercept for the function \(g(x).\)
To determine intercepts from the algebraic expression of the function, notice from the graphs given above that the \(x\)-intercepts are the values of \(x\) which produce \(y = 0\text{.}\) So, to find an \(x\)-intercept (or the intercept on the horizontal axis if the independent variable is called something other than \(x\)) set the function equal to \(0\) and solve for the independent variable.
Determine the \(x\)-intercept(s) of \(f(x) = 3x+2\text{.}\) We set the function equal to \(0\) and solve for \(x:\)
\begin{align*} 3x + 2 & = 0\\ 3x & = -2\\ x & = -2/3 \end{align*}The \(x\)-intercept is the point \((-2/3,0).\) It is at \(x = -2/3.\checkmark\)
Determine the \(y\)-intercept(s) of \(f(x) = 3x+2\text{.}\) This one is easier. Since the \(y\)-intercept corresponds to when \(x=0,\) you just need to find the \(y\)-value when \(x = 0.\) Plug in \(x = 0\) and simplify:
\begin{align*} y & = 3(0) + 2\\ & = 2 \end{align*}The \(y\)-intercept is the point \((0,2).\) It is at \(y = 2.\checkmark\)
We may slightly abuse the notation and say “The \(y\)-intercept is \(2\)” instead of referring to the point \((0,2).\)
Consider the following graph:
This is our earlier “U” shaped graph (\(y = x^2\)) turned on its side. This might seem like a small change. But ask this question: what is \(y\) when \(x = 3\text{?}\) This question has two answers. This graph contains the points \((3,-9)\) and \((3, 9)\text{.}\) So when \(x = 3\text{,}\) \(y\) is both \(9\) and \(-9\) on this graph.
This violates the only restriction on functions-the rule of consistency. Remember that the \(x\)-axis is the independent variable, the \(y\)-axis the dependent. In this case, one “input” value (\(3\)) is leading to two different “output” values (\(-9\text{,}\) \(9\)) We can therefore conclude that this graph does not represent a function at all. No function, no matter how simple or complicated, could produce this graph.
This idea leads us to the “vertical line test,” the graphical analog of the rule of consistency.
The Vertical Line Test: If you can draw any vertical line that touches a graph in two places, then that graph violates the rule of consistency and therefore does not represent any function.
It is important to understand that the vertical line test is not a new rule! It is the graphical version of the rule of consistency. If any vertical line touches a graph in two places, then the graph has two different output values for the same input value, and this is the only thing that functions are not allowed to do.
Suppose the following is the graph of the function \(y = f(x)\text{.}\)
We can see from the graph that the domain of the graph is \(-2 \leq x \leq 4\) and the range is \(-1 \leq y \leq 2\text{.}\)
Question: What does the graph of \(y = f (x) + 2\) look like?
This might seem an impossible question, since we do not even know what the function \(f(x)\) is. But we don't need to know that in order to plot a few points.
| \(x\) | \(f(x)\) | \(f(x)+2\) | so \(y=f(x)\) contains this point | and \(y=f(x)+2\) contains this point |
| \(-2\) | \(2\) | \(4\) | \((-2,2)\) | \((-2,4)\) |
| \(-1\) | \(-1\) | \(1\) | \((-1,-1)\) | \((-1,1)\) |
| \(0\) | \(2\) | \(4\) | \((0,2)\) | \((0,4)\) |
| \(4\) | \(0\) | \(2\) | \((4,0)\) | \((4,2)\) |
If you plot these points on a graph, the pattern should become clear. Each point on the graph is moving up by two. This comes as no surprise: since you added \(2\) to each \(y\)-value, and adding \(2\) to a \(y\)-value moves any point up by \(2\text{.}\) So the new graph will look identical to the old, only moved up by \(2\text{.}\)
In a similar way, it should be obvious that if you subtract \(10\) from a function, the graph moves down by \(10\text{.}\) Note that, in either case, the domain of the function is the same, but the range has changed.
These translations work for any function. Hence, given the graph of the function \(y = \sqrt{x}\) below (which you could generate by plotting points), you can produce the other two graphs without plotting points, simply by moving the first graph up and down.
Adding or subtracting a constant from \(f(x)\text{,}\) as described above, is one example of a vertical translation: it moves the graph up and down. There are other examples of vertical translations.
For instance, what does doubling a function do to a graph? Let's return to our original function:
What does the graph \(y = 2f(x)\) look like? We can make a table similar to the one we made before.
| \(x\) | \(f(x)\) | \(2f(x)\) | so \(y=2f(x)\) contains this point |
| \(-2\) | \(2\) | \(4\) | \((-2,4)\) |
| \(-1\) | \(-1\) | \(-2\) | \((-1,-2)\) |
| \(0\) | \(2\) | \(4\) | \((0,4)\) |
| \(4\) | \(0\) | \(0\) | \((4,0)\) |
In general, the high points move higher; the low points move lower. The entire graph is vertically stretched, with each point moving farther away from the \(x\)-axis.
Similarly, \(y = \frac{1}{2}f(x)\) yields a graph that is vertically compressed, with each point moving toward the \(x\)-axis.
Finally, what does \(y = -f(x)\) look like? All the positive values become negative, and the negative values become positive. So, point by point, the entire graph flips over the \(x\)-axis.
Vertical translations affect the \(y\)-value; that is, the output, or the function itself. Horizontal translations affect the \(x\)-value; that is, the numbers that come in. They often do the opposite of what it naturally seems they should.
Let's return to our original function \(y = f(x)\text{.}\)
Suppose you were asked to graph \(y = f(x + 2)\text{.}\) Note that this is not the same as \(f(x) + 2\text{!}\) The latter is an instruction to run the function, and then add 2 to all results. But \(y = f(x + 2)\) is an instruction to add \(2\) to every \(x\)-value before plugging it into the function.
\(f(x) + 2\) changes the output \(y\text{,}\) and therefore shifts the graph vertically
\(f(x + 2)\) changes the input \(x\text{,}\) and therefore shifts the graph horizontally.
But which way? In analogy to the vertical translations, you might expect that adding two would shift the graph to the right. But let's make a table of values again.
| \(x\) | \(x+2\) | \(f(x+2)\) | so \(y=f(x+2)\) contains this point |
| \(-4\) | \(-2\) | \(f(-2)=2\) | \((-4,2)\) |
| \(-3\) | \(-1\) | \(f(-1)=-1\) | \((-3,-1)\) |
| \(-2\) | \(0\) | \(f(0)=2\) | \((-2,2)\) |
| \(2\) | \(4\) | \(f(4)=0\) | \((2,0)\) |
This is a very subtle, very important point-please follow it closely and carefully! First of all, make sure you understand where all the numbers in that table came from. Then look what happened to the original graph.
NOTE: The original graph \(f(x)\) contains the point \((-2, 2)\text{;}\) therefore, \(f(x + 2)\) contains the point \((-4, 2)\text{.}\) The point has moved two spaces to the left.
You see what I mean when I say horizontal translations “often do the opposite of what it naturally seems they should”? Adding two moves the graph to the left.
Why does it work that way? Here is my favorite way of thinking about it. \(f(x + 2)\) is an instruction that says to each point, “look two spaces to your right, and copy what the original function is doing there.” At \(x = 2\) it does what \(f(x)\) does at \(x = 4\text{.}\) At \(x = -3\text{,}\) it copies \(f(-1)\text{.}\) And so on. Because it is always copying \(f(x)\) from its right, this graph ends up being a copy of \(f(x)\) moved to the left. If you understand this way of looking at it, all the rest of the horizontal translations will make sense.
Of course, as you might expect, subtraction has the opposite effect: \(f(x - 6)\) takes the original graph and moves it \(6\) units to the right. In either case, these horizontal translations affect the domain of the original function, but not its range.
Recall that \(y = 2 f(x)\) vertically stretches a graph; \(y = \frac{1}{2}f(x)\) vertically compresses. Just as with addition and subtraction, we will find that the horizontal equivalents work backward.
| \(x\) | \(2x\) | \(f(2x)\) | so \(y=f(2x)\) contains this point |
| \(-1\) | \(-2\) | \(2\) | \((-1,2)\) |
| \(-\frac{1}{2}\) | \(-1\) | \(-1\) | \(\left(-\frac{1}{2},-1\right)\) |
| \(0\) | \(0\) | \(2\) | \((0,2)\) |
| \(2\) | \(4\) | \(0\) | \((2,0)\) |
The original graph \(f(x)\) contains the point \((4, 0)\text{;}\) therefore, \(f(2x)\) contains the point \((2, 0)\text{.}\) Similarly, \((-1,-1)\) becomes \(\left(-\frac{1}{2},-1\right)\text{.}\) Each point is closer to the \(y\)-axis; the graph has horizontally compressed.
We can explain this the same way we explained \(f(x+2)\text{.}\) In this case, \(f(2x)\) is an instruction that says to each point, “Look outward, at the \(x\)-value that is double yours, and copy what the original function is doing there.” At \(x = 2\) it does what \(f(x)\) does at \(x = 4\text{.}\) At \(x = -\frac{1}{2}\text{,}\) it copies \(f(-1)\text{.}\) And so on. Because it is always copying \(f(x)\) outside itself, this graph ends up being a copy of \(f(x)\) moved inward; ie a compression. Similarly, \(f(\frac{1}{2}x)\) causes each point to look inward toward the y-axis, so it winds up being a horizontally stretched version of the original.
Finally, \(y = f(-x)\) does precisely what you would expect: it flips the graph around the y-axis. \(f(2)\) is the old \(f(-2)\) and so on.
All of these translations do not need to be memorized: only the general principles need to be understood. But once they are properly understood, even a complex graph such as \(y = -2 (x + 3)^2 + 5\) can be graphed easily. You take the (known) graph of \(y = x^2\text{,}\) flip it over the \(x\)-axis (because of the negative sign), stretch it vertically (the \(2\)), move it to the left by \(3\text{,}\) and move it up \(5\text{.}\)
With a good understanding of translations, and a very simple list of known graphs, it becomes possible to graph a wide variety of important functions. To complete our look at translations, let's return to the graph of \(y = \sqrt{x}\) in a variety of flavors.
