Read Functions and Values from Graphs | Grade 8 Math | LevelKite

You already know that a function gives each allowed input exactly one output, while different inputs may share an output. The set of allowed inputs is called the domain. You have used this rule with tables and ordered pairs. For each graph in this lesson, the text states which inputs it covers, and the grid lines are $1$ unit apart.

A graph shows input-output pairs as points. In a point $(x, y)$ , the input $x$ is measured along the horizontal axis and the output $y$ along the vertical axis. Every point with input $2$ , for example, lies on one vertical line: the line through $2$ on the horizontal axis. So each point where that line meets a graph gives one output for input $2$ , and counting those points tells you how many outputs input $2$ has.

Key concept

The vertical-line check

A graph shows $y$ as a function of $x$ when every input in its stated domain has exactly one output. Different inputs may share an output, so two points at the same height are fine when their inputs differ.

To check, imagine a vertical line at every input in the domain:

If each of these lines meets the graph exactly once, the graph is a function on that domain.
If any of these lines meets the graph at two or more points, that input has two or more outputs, so the graph is not a function of $x$ .

Apply the check to the whole graph across its whole domain. Testing a few sample inputs leaves the rest untested, but one failing input is enough to rule out a function.

Circle with center

(2, 1)

and radius

3

, passing through

(2, 4)

and

(2, -2)

, two points with the same input

2

.

Worked example

Testing a circle

The figure shows a circle with center $(2, 1)$ and radius $3$ , so every point on it is exactly $3$ units from the center. Its inputs run from $2 - 3 = -1$ to $2 + 3 = 5$ . Is the circle a function of $x$ ?

Choose an input from −1 to 5 to test. The center's input, x = 2, is a good choice, because the vertical line there passes right through the center.
$x = 2$
Move 3 units straight up and straight down from the center (2, 1). Each new point is exactly 3 units from the center, so it lies on the circle. Moving straight up or down changes only y, so both points keep the input 2.
$(2, 1 + 3) = (2, 4) \quad \text{and} \quad (2, 1 - 3) = (2, -2)$
So input 2 has two outputs, 4 and −2. One input with two outputs breaks the rule, so you can stop here: no other input needs checking.

Answer

The circle is not a function of $x$ , because the input $x = 2$ has two outputs, $y = 4$ and $y = -2$ .

The curve

x = y^2 - 4y + 5

for

0 \le y \le 4

, which passes through both

(2, 1)

and

(2, 3)

.

A parabola is a U-shaped curve. The curve above is a sideways parabola: its equation $x = y^2 - 4y + 5$ , for $0 \le y \le 4$ , finds $x$ from $y$ , so the U opens to the right. Putting in values of $y$ gives points on it:

$y = 0$ : $x = 0^2 - 4(0) + 5 = 0 - 0 + 5 = 5$ , one end
$y = 1$ : $x = 1^2 - 4(1) + 5 = 1 - 4 + 5 = 2$
$y = 2$ : $x = 2^2 - 4(2) + 5 = 4 - 8 + 5 = 1$ , the turning point, where the curve changes direction
$y = 3$ : $x = 3^2 - 4(3) + 5 = 9 - 12 + 5 = 2$
$y = 4$ : $x = 4^2 - 4(4) + 5 = 16 - 16 + 5 = 5$ , the other end

Its inputs run from $1$ , at the turning point, to $5$ , at the ends. Input $2$ appears twice: $(2, 1)$ and $(2, 3)$ are both on the curve, so input $2$ has two outputs. The sideways parabola is not a function of $x$ . The next graph passes the check.

One connected path over the domain

0 \le x \le 8

: a segment, a curve and a segment, joined at

(2, 5)

and

(6, 5)

.

Worked example

Testing a connected path

This graph is one connected path over the domain $0 \le x \le 8$ , with its pieces joined end to end. From left to right, the pieces are the segment from $(0, 1)$ to $(2, 5)$ , the curve $y = x^2 - 8x + 17$ for $2 \le x \le 6$ , and the segment from $(6, 5)$ to $(8, 3)$ . Is the graph a function of $x$ on its domain?

Check that the pieces cover the whole domain. They run over inputs 0 to 2, then 2 to 6, then 6 to 8, so every input from 0 to 8 lies on at least one piece.
$0 \le x \le 2, \quad 2 \le x \le 6, \quad 6 \le x \le 8$
Check inside each piece. Each segment's endpoints have different inputs, so it is not vertical, and a vertical line meets it at most once. Unlike the sideways parabola, the curve's equation finds y by calculating with x, and one calculation gives one result. So inside any one piece, each input has one output. This comes from the stated endpoints and equation, not from how the picture looks.
Check input 2, where the first segment meets the curve, because it belongs to both pieces. The segment ends at (2, 5), and the curve gives the same output, so input 2 still has only one output.
$2^2 - 8(2) + 17 = 4 - 16 + 17 = 5$
Check input 6 the same way. The curve gives 5, which matches the start (6, 5) of the last segment.
$6^2 - 8(6) + 17 = 36 - 48 + 17 = 5$
Notice that inputs 2 and 6 share the output 5. That is allowed: the rule limits how many outputs one input has, not how many inputs share an output.

Answer

Yes. Every input in the domain $0 \le x \le 8$ has exactly one output, so the graph is a function of $x$ on that domain.

Because a function gives each input in its domain exactly one output, though different inputs may share an output, you can ask for the output of an input. To read it, start at the input on the horizontal axis and look straight up to the graph (straight down if the graph is below the axis). Then look straight across to the vertical axis: the number there is the output. This works because the point you reach has your input as its $x$ -coordinate, and its height is its $y$ -coordinate, the output.

Read values only at marked inputs, the inputs where the figure marks a point on the graph. They sit on grid lines, and the figure gives each marked point exactly. Between marks, the drawing can only suggest an estimate, not an exact value. After reading a value, check it against the piece it lies on, using that piece's equation or its endpoints. An input outside the domain has no output, because the vertical line there misses the graph.

The same path with inputs

1

,

3

and

7

marked; the domain is still

0 \le x \le 8

.

Worked example

Reading values from the path

Use the marked points on this graph, whose domain is $0 \le x \le 8$ , to find the outputs for inputs $1$ , $3$ and $7$ , and check each one against its piece. Then decide whether input $9$ has an output.

Input 1: start at 1 on the horizontal axis, look up to the marked point, then look across to the vertical axis, which gives 3. Check it: input 1 is on the segment from (0, 1) to (2, 5). On a straight segment, equal steps in input give equal steps in output, so input 1, halfway from 0 to 2, has the output halfway from 1 to 5.
$\frac{1 + 5}{2} = \frac{6}{2} = 3$
Input 3: look up from 3 to the marked point on the curve, then across to the output 2. Check it by putting x = 3 into the curve's equation, which covers inputs 2 to 6.
$y = 3^2 - 8(3) + 17 = 9 - 24 + 17 = 2$
Input 7: look up from 7 to the marked point, then across to the output 4. Check it: input 7 is halfway along the segment from (6, 5) to (8, 3), so its output is halfway from 5 to 3.
$\frac{5 + 3}{2} = \frac{8}{2} = 4$
Input 9 is greater than 8, so it lies outside the domain. The vertical line at 9 misses the graph, so input 9 has no output here. Do not extend the last segment to guess one.
$9 > 8$

Answer

Input $1$ has output $3$ , input $3$ has output $2$ , and input $7$ has output $4$ . Input $9$ is outside the domain $0 \le x \le 8$ , so the graph gives it no output.

A graph is a function of $x$ when every input in its stated domain has exactly one output: each vertical line at an input in the domain meets the graph exactly once, though different inputs may share an output. A circle and a sideways U-shaped parabola each have an input with two outputs, so neither is a function of $x$ . On a function's graph, find an output by looking from a marked input up (or down) to the graph and across, then confirm it with the piece's equation or endpoints. Inputs outside the domain have no output. Next, you will find the domain and range of a function, where the range is the set of all its outputs.