Evaluate a linear function and complete its table | Grade 8 Math | LevelKite

A function is a rule that gives each permitted input exactly one output. The set of permitted inputs is the function's domain. Two different inputs may share an output, but one input never has two outputs. In this lesson you will use a rule in both directions. You will go forward from an input to its output. You will also go backward from an output to an input that produces it, when such an input exists.

A linear function has a rule of the form $f(x) = mx + b$ , where $m$ and $b$ are fixed numbers. This is the slope-intercept form you already know, with $f(x)$ written in place of $y$ . Read $f(x)$ as f of x. It names the output for the input $x$ , and it does not mean $f$ times $x$ . For example, $f(x) = -2x + 5$ has $m = -2$ and $b = 5$ . No situation limits this rule, and it works for every number, so its domain is all real numbers.

Evaluating a function at an input means finding the output for that input. To do it, you substitute: you replace $x$ with the input and then do the arithmetic, multiplying before you add or subtract. This works because the rule tells you exactly what to do with any input in the domain. Put the input in parentheses when you substitute. That keeps a negative sign or a fraction attached to the number being multiplied.

Worked example

Evaluate at a negative input and a fraction

Use $f(x) = -2x + 5$ , whose domain is all real numbers. Evaluate $f(-3)$ and $f\left(\tfrac{1}{2}\right)$ .

Replace x with −3 and write it in parentheses, so the rule still reads −2 times the input.
$f(-3) = -2(-3) + 5$
Multiply first. A negative times a negative is positive, so −2 times −3 is 6. Then add 5.
$f(-3) = 6 + 5 = 11$
Replace x with 1/2, again in parentheses, so the whole fraction is what gets multiplied by −2.
$f\left(\tfrac{1}{2}\right) = -2\left(\tfrac{1}{2}\right) + 5$
Multiply first. −2 times 1/2 is half of −2, which is −1. Then add 5.
$f\left(\tfrac{1}{2}\right) = -1 + 5 = 4$

Answer

The output for input $-3$ is $11$ , so $f(-3) = 11$ . The output for input $\tfrac{1}{2}$ is $4$ , so $f\left(\tfrac{1}{2}\right) = 4$ .

A table for a function lists input–output pairs, with one pair in each row. When the rule is given, any row you fill in must make the rule true, so the rule decides each missing entry. (A table with no rule tells you only the pairs it shows.) To complete a table, you find every missing entry. Some rows give the input and are missing the output. Other rows give the output and are missing the input.

Key concept

One rule, two directions

For a linear function $f(x) = mx + b$ , the rule settles every missing entry of its table.

Input given: substitute the permitted input into the rule to get its output.
Output given: if $m \ne 0$ , set $mx + b$ equal to the output and solve for $x$ . There is exactly one solution. Keep it only if it is in the domain. If it is not in the domain, no permitted input has that output.

If $m = 0$ , every input gives the output $b$ . In that case an output other than $b$ has no input, and the output $b$ comes from every input in the domain.

f(x) = mx + b

Now a situation limits the domain. A pump empties a water tank. Let $t$ be the time in minutes since the pump started, and let $W(t)$ be the liters of water left at time $t$ . The rule is $W(t) = 120 - 15t$ . The pump runs from $t = 0$ until the tank is empty, and the rule describes the tank only during that time. So you first find when the tank is empty. Here $m = -15$ , which is not $0$ . Solving for a missing time therefore always gives exactly one value of $t$ , which you then check against the domain.

Water left in the tank. Each ? marks a missing entry.

Row	Time t, in minutes	Water left W(t), in liters
1	$2.5$	?
2	?	$30$
3	?	$50$
4	?	$150$

Worked example

Find the domain, then complete rows 1 and 2

For $W(t) = 120 - 15t$ , find the domain. Then complete rows 1 and 2 of the table: find the water left at $t = 2.5$ , and find the time when $30$ liters are left.

The tank is empty when the output is 0, so set the rule equal to 0. Add 15t to both sides. Doing the same thing to both sides keeps them equal.
$120 - 15t = 0 \;\Rightarrow\; 120 = 15t$
Divide both sides by 15, which is allowed because 15 is not 0. So the tank is empty at 8 minutes. The rule describes the tank from the start at 0 minutes until then. The domain is therefore every time from 0 to 8 minutes, including both ends.
$t = \frac{120}{15} = 8, \qquad \text{domain: } 0 \le t \le 8$
Row 1 gives the input 2.5, which is between 0 and 8, so it is permitted. Substitute it in parentheses, and multiply before subtracting.
$W(2.5) = 120 - 15(2.5) = 120 - 37.5 = 82.5$
Row 2 gives the output 30, so set the rule equal to 30. Subtract 120 from both sides, which keeps them equal.
$120 - 15t = 30 \;\Rightarrow\; -15t = -90$
Divide both sides by −15, which is allowed because −15 is not 0. A negative divided by a negative is positive.
$t = \frac{-90}{-15} = 6$
Check the domain. 6 is between 0 and 8, so it is a permitted input. Then check the row by substituting 6 back into the rule.
$W(6) = 120 - 15(6) = 120 - 90 = 30$

Answer

The domain is $0 \le t \le 8$ . After $2.5$ minutes, $82.5$ liters are left, so $W(2.5) = 82.5$ . Exactly $30$ liters are left at $t = 6$ minutes.

Worked example

Complete rows 3 and 4

Complete rows 3 and 4 of the table: find the time when $50$ liters are left and the time when $150$ liters are left. Use the domain $0 \le t \le 8$ that you found.

Row 3 gives the output 50, so set the rule equal to 50. Subtract 120 from both sides. Then divide both sides by −15, which is not 0; a negative divided by a negative is positive. Simplify the fraction by dividing top and bottom by 5.
$120 - 15t = 50 \;\Rightarrow\; -15t = -70 \;\Rightarrow\; t = \frac{-70}{-15} = \frac{70}{15} = \frac{14}{3}$
The exact time is 14/3 minutes, which is 4 and 2/3 minutes. That equals 4 minutes 40 seconds, because 2/3 of 60 seconds is 40 seconds. Dividing 14 by 3 gives 4.666…, so 4.67 is only an approximation. The time is between 0 and 8, so it is permitted. Check the row.
$W\left(\tfrac{14}{3}\right) = 120 - 15\left(\tfrac{14}{3}\right) = 120 - \tfrac{210}{3} = 120 - 70 = 50$
Row 4 gives the output 150. Solve the same way: subtract 120 from both sides, then divide both sides by −15, which is not 0. A positive divided by a negative is negative.
$120 - 15t = 150 \;\Rightarrow\; -15t = 30 \;\Rightarrow\; t = \frac{30}{-15} = -2$
Check the domain. −2 is less than 0, so it is not permitted. It would mean 2 minutes before the pump started, which the rule does not describe. Because m = −15 is not 0, t = −2 is the only solution. So no permitted time gives 150 liters.
$-2 < 0$

Answer

Exactly $50$ liters are left at $t = \tfrac{14}{3}$ minutes, which is $4$ minutes $40$ seconds. No permitted input has output $150$ : the only solution, $t = -2$ , is outside the domain, so the tank never holds $150$ liters while the pump runs.

Takeaway: One rule works in both directions. Substitute a permitted input to get its one output. To work backward from an output, set the rule equal to that output and solve. When $m \ne 0$ , you get exactly one possible input, and it counts only if the domain permits it. Next, you will start from a table or graph instead of a rule. You will find the rate of change and the initial value, then write the linear function.