Precalculus-I · Unit 1 · Video 4 · Interactive Practice

Evaluating and Solving from Tables and Graphs

IKey Formulas

Question	Given	From a table	From a graph
Evaluate $f(a)$	the input $a$	find $a$ among the inputs, read the output paired with it	go along the vertical line $x = a$ to the curve; $f(a)$ is the height there
Solve $f(x) = b$	the output $b$	find every place $b$ appears among the outputs, read each paired input	go along the horizontal line $y = b$ ; each point where it meets the curve gives one solution

Key Insight: evaluating at an input in the domain gives exactly one output, while solving can give one input, several inputs, or none.

IIEvaluating and Solving from a Table

Evaluating reads from an input down to its one output; solving reads back from an output to every input.

Question Input n 3

IIIEvaluating from a Graph

To evaluate $f(a)$ , go up from $a$ on the input axis to the curve: the output is the height there, not the point.

Input a 2

IVSolving from a Graph

Every point where the line $y = b$ meets the curve gives one solution of $f(x) = b$ .

Output b 4

💡 The next video turns these two lines into tests: the vertical line test for whether a graph is a function, and the horizontal line test for whether a function is one-to-one.

VQuiz Questions

Problem 1 · Evaluate from a Table

Given: the function $h$ in the table — find $h(4)$ .

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$h(x)$	$3$	$9$	$4$	$9$	$6$	$7$

✅ Correct! The input

4

sits above the output

6

, so

h(4) = 6

❌ Not quite.

2

is the input whose output is

4

— that answers

h(x) = 4

. In

h(4)

the

4

is the input.

❌ Not quite.

9

sits under

x = 3

, one column to the left: that is

h(3)

❌ Not quite.

7

sits under

x = 5

, one column to the right: that is

h(5)

❌ Not quite. Find

4

in the

x

row and read the entry directly below it.

Show solution

Evaluating means the input is known. Find $x = 4$ in the input row; directly below it is $6$ .

h(4) = 6

Only one output sits under an input, so there is exactly one answer.

Problem 2 · Solve from a Table

Given: the same table of $h$ — solve $h(x) = 9$ .

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$h(x)$	$3$	$9$	$4$	$9$	$6$	$7$

✅ Correct!

9

appears twice in the output row, so there are two solutions:

h(1) = 9

and

h(3) = 9

❌ Close, but incomplete.

h(1) = 9

is right, but the scan stopped at the first match. Keep reading the output row.

❌ Close, but incomplete.

h(3) = 9

is right, but

9

also appears earlier in the output row.

❌ Not quite.

9

is not an input, but here it is an output: look for

9

in the

h(x)

row, not the

x

row.

❌ Not quite. Find every

9

in the output row and read the input above each one.

Show solution

Solving means the output is known. Scan the $h(x)$ row for $9$ : it appears under $x = 1$ and again under $x = 3$ , and nowhere else.

h(x) = 9 \iff x = 1 \text{ or } x = 3

Check: $h(1) = 9$ and $h(3) = 9$ .

Problem 3 · Evaluate, Then Solve, on the Parabola

Given: the graph of $f$ from Sections III and IV, an upward parabola with vertex $(1, 0)$ passing through $(0, 1)$ , $(2, 1)$ , $(-1, 4)$ and $(3, 4)$ — find $f(0)$ , then solve $f(x) = f(0)$ .

What is $f(0)$ ?

Solve $f(x) = f(0)$ .

✅ Correct!

f(0) = 1

, and the line

y = 1

meets the parabola twice, above

x = 0

and

x = 2

❌ Not quite.

0

is the input. Go up from

x = 0

to the curve: the curve is at height

1

there.

❌ Close, but…

(0, 1)

is the point on the graph. The output is its height, the

y

-coordinate, a single number.

❌ Not quite.

2

is the other input with height

1

, not an output. Read the height above

x = 0

❌ Close, but incomplete.

x = 0

works, but the line

y = 1

crosses the parabola a second time on the other side of the vertex.

❌ Not quite.

1

is the output

b

you are solving for, not an input. Also

f(1) = 0

, not

1

❌ Close, but incomplete.

x = 2

works, but so does the input you started from:

f(0) = 1

as well.

❌ Check $f(0)$ . Go up the vertical line

x = 0

to the curve and read the height.

❌ Check the solutions. Draw

y = 1

and read the input below every crossing.

Show solution

Step 1: Evaluate. Go up the vertical line $x = 0$ to the curve. It meets the parabola at $(0, 1)$ , whose height is $1$ , so $f(0) = 1$ .

Step 2: Solve $f(x) = 1$ . Go across the horizontal line $y = 1$ . It meets the parabola at $(0, 1)$ and at $(2, 1)$ . Drop down from each point to the input axis:

f(x) = 1 \iff x = 0 \text{ or } x = 2

Check: $f(0) = 1$ and $f(2) = 1$ . The input you started from is always one of the solutions; the parabola supplies a second.

Problem 4 · Both Directions on a New Graph

Given: a function $k$ is known only from its graph. The vertical line $x = 3$ meets the graph at $(3, 5)$ . The horizontal line $y = 3$ meets the graph at exactly three points: $(-2, 3)$ , $(1, 3)$ and $(4, 3)$ — find $k(3)$ and solve $k(x) = 3$ .

What is $k(3)$ ?

Solve $k(x) = 3$ .

✅ Correct! The vertical line gives the one output

k(3) = 5

; the horizontal line gives three solutions of

k(x) = 3

❌ Not quite. The

3

k(3)

is an input. The vertical line

x = 3

meets the graph at height

5

❌ Close, but…

(3, 5)

is the point; the output is its height,

5

❌ Not quite.

k(4) = 3

, so

4

answers the solving question

k(x) = 3

, not the evaluating question

k(3)

❌ Not quite.

5

is the output

k(3)

from the vertical line. Solving

k(x) = 3

uses the horizontal line

y = 3

❌ Close, but incomplete.

k(1) = 3

is true, but the line

y = 3

meets the graph at three points, and each gives a solution.

❌ Not quite. Here

3

is the output.

k(3) = 5

, not

3

, so

x = 3

is not a solution.

❌ Check $k(3)$ . Use the vertical line

x = 3

and read the height where it meets the graph.

❌ Check the solutions. Use the horizontal line

y = 3

and read the input of every meeting point.

Show solution

Evaluate: the input $3$ is known, so use the vertical line $x = 3$ . It meets the graph at $(3, 5)$ , whose height is $5$ : $k(3) = 5$ .

Solve: the output $3$ is known, so use the horizontal line $y = 3$ . It meets the graph at $(-2, 3)$ , $(1, 3)$ and $(4, 3)$ . The input of each meeting point is a solution:

k(x) = 3 \iff x = -2,\ x = 1 \text{ or } x = 4

The two questions use different numbers $3$ in different roles: in $k(3)$ the $3$ is an input, in $k(x) = 3$ it is an output.

Solved: 0 / 4