Graphs of Functions

The table is $(-4,-1)$, $(-2,0)$, $(0,1)$, $(2,2)$, $(4,3)$. The graph meets the $x$-axis at $(-2,0)$ and the $y$-axis at $(0,1)$.

Step 1: the table. Substitute each chosen $x$ into the rule.

Step 2: plot the five points $(x,f(x))$ and join them.

Step 3: read off the meetings with the axes.

The graph meets the $x$-axis where $y=0$. From the table that happens at $x=-2$, so the point is $(-2,0)$.

The graph meets the $y$-axis where $x=0$. From the table $f(0)=1$, so the point is $(0,1)$.

Check by algebra. Setting $f(x)=0$ gives ${\displaystyle \frac{x}{2}}+1=0$, so ${\displaystyle \frac{x}{2}}=-1$ and $x=-2$ — matching the picture. Setting $x=0$ gives $y=1$ — also matching.

Notice that the five plotted points lie on one straight line, which is why joining them with a ruler is legitimate here. Do not assume that this happens for every function; the next exercise shows a case where it does not.

Table: $-3\mapsto 5$, $-2\mapsto 0$, $-1\mapsto -3$, $0\mapsto -4$, $1\mapsto -3$, $2\mapsto 0$, $3\mapsto 5$.   (a) at $(-2,0)$ and $(2,0)$   (b) $x=-3$ and $x=3$

Step 1: the table. Square first, then subtract $4$. Remember that squaring a negative number gives a positive result.

Step 2: plot and join with a smooth curve.

Here the points clearly do not lie on a line, so they are joined by a smooth free-hand curve as the section describes.

(a) The graph meets the $x$-axis where $y=0$. From the table this happens at $x=-2$ and at $x=2$, so the points are $(-2,0)$ and $(2,0)$.

Confirmation by algebra: $x^2-4=0$ gives $(x-2)(x+2)=0$, so $x=2$ or $x=-2$.

(b) To find where $f(x)=5$, draw the horizontal line $y=5$ (dashed above) and see where it cuts the curve. It cuts at the two points $(-3,5)$ and $(3,5)$, so

Confirmation by algebra: $x^2-4=5$ gives $x^2=9$, so $x=3$ or $x=-3$. Both answers must be given; keeping only $x=3$ would miss half the picture, and the graph makes that obvious.

A "V" shape with its lowest point at $(0,-1)$, passing through $(-1,0)$ and $(1,0)$.

Step 1: the table. The absolute value of a number is its size without regard to sign, so $|-3|=3$, $|-2|=2$, and so on.

Step 2: plot and join.

Reading the picture. The plotted points fall exactly on two straight pieces that meet at $(0,-1)$, producing a sharp corner rather than a smooth turn. This is not a mistake in the plotting: for $x\ge 0$ we have $|x|=x$, so the outputs are $x-1$, and for $x\le 0$ we have $|x|=-x$, so the outputs are $-x-1$. Each side is a straight line, and the two lines meet at the origin's neighbour $(0,-1)$.

Check. $f(-2.5)=2.5-1=1.5$, which is halfway up the left branch between the plotted points $(-3,2)$ and $(-2,1)$, as the straight edge suggests. ✔

A warning: when a graph has a corner, joining the points with a rounded curve would misrepresent the function. Compute a few extra values near the suspicious spot before deciding how to join.

The points are $(0,0)$, $(0.25,0.5)$, $(1,1)$, $(4,2)$, $(9,3)$, joined by a curve that rises steeply at first and then flattens.

Step 1: the table. Choosing perfect squares makes the arithmetic exact.

Step 2: plot and join with a smooth curve.

Two features worth noticing.

Nothing to the left of $0$. The natural domain of $\sqrt{x}$ is $x\ge 0$, so the graph simply stops at the origin. There is no part of the picture with $x$ negative — not because we ran out of paper, but because $f$ has no output there.

Only nonnegative heights. The symbol $\sqrt{}$ means the nonnegative square root, so every plotted point sits on or above the $x$-axis. The point $(4,-2)$ is not on this graph, even though $(-2{)}^2=4$.

Check. The value halfway along, $f(2.25)=1.5$, since ${1.5}^2=2.25$. That point lies neatly on the curve drawn. ✔

Two separate branches, one in the region where and , one where and . There is no point at $x=0$, because $f(0)$ is undefined.

Step 1: the table. Recall that ${\displaystyle \frac{1}{1/2}}=2$ and that the reciprocal of a negative number is negative.

Step 2: plot the points.

What happens at $x=0$. The number $f(0)={\displaystyle \frac{1}{0}}$ does not exist, so $0$ is not in the natural domain and the graph has no point at all on the $y$-axis. The picture consists of two separate branches, one on each side of the $y$-axis.

Why the two branches must not be joined. A curve drawn from the left branch to the right branch would have to cross the $y$-axis, and any point it placed there would be a point of the graph with $x=0$ — which cannot exist. Joining the pieces would therefore draw something that is not the graph of $f$.

Check. As the inputs get closer to $0$ from the right ($x=1$, then $\frac{1}{2}$, then $\frac{1}{4}$) the outputs $1$, $2$, $4$ grow without bound, matching the steep rise of the right branch. As the inputs grow ($x=1,2,4$) the outputs $1,0.5,0.25$ shrink toward $0$ without ever reaching it, matching the flattening tail.

(a) $f(0)=-1$, $f(1)=0$, $f(2)=3$   (b) $x=-1$ and $x=1$   (c) $x=-2$ and $x=2$

How to read a value from a graph. To find $f(a)$, start at $a$ on the horizontal axis, move straight up or down until you meet the curve, and read the height of that meeting point on the vertical axis. The definition of the graph guarantees this works, since the graph consists precisely of the points $(x,f(x))$.

(a)

• At $x=0$: moving down from $0$ we meet the curve at its lowest point, whose height is $-1$. So $f(0)=-1$.
• At $x=1$: moving up from $1$ we meet the curve exactly on the horizontal axis, at height $0$. So $f(1)=0$.
• At $x=2$: moving up from $2$ we meet the curve at height $3$. So $f(2)=3$.

(b) Now the question runs the other way: we are given the output $0$ and must find the inputs. Outputs of $0$ are heights of $0$, that is, points on the horizontal axis. The curve meets that axis at $(-1,0)$ and $(1,0)$, so

(c) Draw (or imagine) the horizontal line at height $3$ and see where it meets the curve. It meets it at $(-2,3)$ and $(2,3)$, so

A point worth stressing. In part (a) each question had exactly one answer, whereas in parts (b) and (c) there were two. Going up from a point of the horizontal axis can meet the curve only once, since $f$ assigns exactly one output to each input. Going across from a point of the vertical axis may meet the curve many times, because different inputs are allowed to share an output.

$(0,1)$, $(2,-1)$ and $(-1,5)$ lie on the graph; $(3,2)$ does not.

A point $(a,b)$ lies on the graph of $f$ exactly when $b=f(a)$ — that is, when the second coordinate is the output produced by the first. So compute $f$ at each first coordinate and compare.

$(0,1)$:

The required second coordinate is $1$, and $1=1$. The point is on the graph.

$(2,-1)$:

The required second coordinate is $-1$, and $-1=-1$. The point is on the graph.

$(3,2)$:

But the point offers the second coordinate $2$, and $1\ne 2$. So $(3,2)$ is not on the graph. (The point $(3,1)$ is.)

$(-1,5)$:

Since $5=5$, the point is on the graph. Watch the two sign changes here: $(-1{)}^2=+1$ and $-3(-1)=+3$.

Summary. Three of the four points lie on the graph. Note that testing a point requires only one substitution — there is no need to sketch anything.

(a) $a=2$   (b) $(-3,17)$   (c) $(-{\displaystyle \frac{1}{\sqrt{2}}},0)$ and $({\displaystyle \frac{1}{\sqrt{2}}},0)$

(a) Saying that $(2,7)$ is on the graph means $f(2)=7$. Substitute $x=2$ into the rule:

Since $(2{)}^2=4$, this is

Add $1$ to both sides:

Divide by $4$:

So the function is $f(x)=2x^2-1$.

Check. $f(2)=2(4)-1=8-1=7$. ✔

(b) A point of the graph with first coordinate $-3$ has second coordinate $f(-3)$:

The point is $(-3,17)$. Note $(-3{)}^2=9$, not $-9$.

(c) A point lies on the $x$-axis exactly when its second coordinate is $0$, so solve $f(x)=0$:

Add $1$ to both sides:

Divide by $2$:

Take square roots, remembering both signs:

Rationalising, ${\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{\sqrt{2}}{2}}\approx 0.707$. So the graph meets the $x$-axis at $(-{\displaystyle \frac{\sqrt{2}}{2}},0)$ and $({\displaystyle \frac{\sqrt{2}}{2}},0)$.

Check. $f\left({\displaystyle \frac{1}{\sqrt{2}}}\right)=2\cdot {\displaystyle \frac{1}{2}}-1=1-1=0$. ✔

$\mathrm{Dom}(g)=[-1,3]$ and $\mathrm{Rng}(g)=[0,2]$.

The domain: project onto the $x$-axis. The domain is the set of $x$-values for which the graph has a point above or below them. The drawn curve starts at the point $(-1,0)$ and ends at the point $(3,2)$, with no gaps in between. Sliding a vertical line across the picture, it meets the curve for every $x$ from $-1$ to $3$ and for no other $x$. Hence

Both endpoints are included because the curve genuinely reaches the solid dots at $(-1,0)$ and $(3,2)$.

The range: project onto the $y$-axis. The range is the set of heights that occur on the curve. The lowest point of the curve is $(-1,0)$, at height $0$; the highest is $(3,2)$, at height $2$. The curve climbs from one to the other without breaks, so every height between $0$ and $2$ occurs somewhere. Hence

Testing individual values, as the section suggests.

• Is $2$ in the domain? Draw the vertical line $x=2$: it hits the curve (at about height $1.74$). Yes.
• Is $4$ in the domain? The vertical line $x=4$ misses the curve entirely. No.
• Is $1$ in the range? Draw the horizontal line $y=1$: it hits the curve at $(0,1)$. Yes.
• Is $2.5$ in the range? The horizontal line $y=2.5$ passes above everything drawn. No.

A caution. The domain is read across the picture and the range up it. Interchanging the two is the most common error, so it is worth naming the axis explicitly each time: domain — $x$-axis; range — $y$-axis.

(i) $y=x^2$ is Graph A; (ii) $y=|x|$ is Graph C; (iii) $y=2x$ is Graph B.

The safest way to match is to compute two or three points from each formula and see which picture contains them.

(iii) $y=2x$. A short table: $x=-1$ gives $y=-2$; $x=0$ gives $y=0$; $x=1$ gives $y=2$. So the graph must have points below the $x$-axis on the left. Only Graph B dips below the axis, so (iii) is Graph B.

(i) $y=x^2$ and (ii) $y=|x|$. Both are never negative, matching Graphs A and C, which both sit on or above the axis and both touch the origin. To separate them, use a value between $0$ and $1$, where squaring and absolute value behave very differently.

Take $x=\frac{1}{2}$:

So the graph of $y=x^2$ must pass below the point $(\frac{1}{2},\frac{1}{2})$, sagging toward the axis near the origin, while the graph of $y=|x|$ must pass exactly through it.

Graph A sags near the origin and leaves it smoothly; Graph C rises in two straight edges meeting in a sharp corner. Straight edges are what $y=|x|$ produces, because for $x\ge 0$ it equals $x$ and for $x\le 0$ it equals $-x$, and each of those is a line.

Hence (i) is Graph A and (ii) is Graph C.

Check with a second value. At $x=2$: $x^2=4$ but $|x|=2$. So beyond $x=1$ the squaring graph climbs much faster than the absolute-value graph, which agrees with the shapes drawn.

The error. The point $(0,0)$ can be on the graph only if $f(0)=0$. But

which is not a real number: division by zero is undefined. So $0$ is not even in the natural domain of $f$, and the graph has no point whatsoever on the $y$-axis. The student has drawn a curve through a place where the function does not exist.

Where the reasoning went wrong. Joining plotted points with a smooth curve is a guess about what happens between them, and it is only a reasonable guess where the function is defined and has no break. The student extended the curve into a region that was never sampled and where the function is undefined.

What really happens near $x=0$. Compute outputs for inputs shrinking toward $0$ from the right:

Far from settling down toward $0$, the outputs grow without bound. The right-hand branch of the graph shoots upward and never crosses the $y$-axis.

On the left of $0$ the outputs are negative and behave in the mirror manner:

So the graph has two separate branches, and there is a gap at $x=0$ that must not be bridged.

The general lesson. Before joining plotted points, check the natural domain. Wherever a value is excluded from the domain, the graph must be interrupted there.

What the two axes mean for a graph. By definition, the graph of $f$ is the set of points $(x,f(x))$ for $x$ in the domain. So:

• a point of the graph lies on the $x$-axis when its second coordinate is $0$, that is, when $f(x)=0$;
• a point of the graph lies on the $y$-axis when its first coordinate is $0$, that is, when the input is $0$.

Why many crossings of the $x$-axis are possible. Nothing in the definition of a function forbids two different inputs from producing the same output, so many inputs may produce the output $0$. For example, $f(x)=x^2-4$ has $f(-2)=0$ and $f(2)=0$, giving two crossings; and a function can be arranged to give the output $0$ at as many inputs as we please. Indeed the constant function $f(x)=0$ produces a graph lying entirely along the $x$-axis.

Why at most one crossing of the $y$-axis. A point of the graph on the $y$-axis must come from the input $0$. If there were two such points, say $(0,p)$ and $(0,q)$ with $p\ne q$, then the single input $0$ would have two different outputs $p$ and $q$. That contradicts the defining property of a function, which allows each input exactly one output. So there cannot be two.

Why "at most" and not "exactly". There may be no crossing at all, namely when $0$ is not in the domain. The graph of $f(x)={\displaystyle \frac{1}{x}}$ is an example: it never touches the $y$-axis, because $f(0)$ does not exist. Similarly $f(x)=\sqrt{x-1}$, whose domain is $[1,\mathrm{\infty })$, has no point above or below $0$.

Summary. Horizontal position is the input, and inputs may not be reused with different outputs; vertical position is the output, and outputs may be repeated freely. The asymmetry of the two axes in this question is exactly the asymmetry built into the definition of a function.

It consists of the five isolated points $(-2,-1)$, $(-1,1)$, $(0,3)$, $(1,5)$, $(2,7)$ — not a line.

Step 1: compute the outputs.

Step 2: the graph is the set of pairs $(x,f(x))$ for $x$ in the domain, and the domain has only five members. So the graph consists of exactly five points:

Step 3: the comparison. With its natural domain, the formula $2x+3$ accepts every real number, so the graph contains a point for every $x$ — for instance $(0.5,4)$ and $(-1.3,0.4)$ — and those points fill out an unbroken straight line.

With the declared domain $\{-2,-1,0,1,2\}$ there is no point above $x=0.5$, because $0.5$ is not an allowable input. The graph is five separated dots, and joining them would add points that do not belong to the function.

The moral. A graph is determined by the rule and the domain. Two functions given by the same formula but different domains have different graphs. This is why a domain restriction must always be stated, as the section on natural domains insisted.

Check. Every dot satisfies $y=2x+3$: for the dot $(1,5)$, $2(1)+3=5$. ✔ And the five first coordinates are exactly the five members of the declared domain. ✔