f(a): up from x = a, then across.
The input lives on the x-axis. The answer lives on the y-axis. The curve is the bridge.
f(3) = 1
up to the curve, across to the axis
f(−4) = −2, f(−2) = 2
endpoints and corners read the same way
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- Reading f(x) from a graph
Reading f(x) from a Graph
Up from x to the curve, across to the axis — that one trip evaluates f. Run it backwards and you solve f(x) = k, solutions counted on sight.
Read the graph
What is f(3)?
Start at x = 3 on the horizontal axis, go straight up to the curve, then across to the vertical axis: you land at height 1. A hands back the input as the output; starting at y = 3 and going across answers a different question — that is solving f(x) = 3.
Solve f(x) = k: start at y = k and come down.
Same trip, backwards — and now there can be several answers.
f(x) = 2 → x = −2 or x = 4
drop down from each meeting point
A touch counts. The line only grazes the peak at (−2, 2), but grazing is still meeting — that x is still a solution.
The trap: f(2) is not f(x) = 2
Asked for f(2), a hurried reader goes to 2 on the y-axis, runs across, and answers −2 or 4 — the solutions of f(x) = 2. But f(2) starts on the x-axis: up from x = 2 the curve is at height 0, so f(2) = 0. A number inside the brackets is an input (start on x); a number after the equals sign is an output (start on y).
Zeros are just f(x) = 0 — the x-axis crossings.
f(x) = 0 at x = −3, 0, 2
three crossings, three zeros
"Zeros", "roots", "x-intercepts" and "solutions of f(x) = 0" are four names for the same dots. Exams rotate the vocabulary hoping one version throws you.
Count solutions by sliding the horizontal line.
How many times does y = k cut the curve? Slide it up and down and watch the count change.
| Line | Meets the curve | Solutions of f(x) = k |
|---|---|---|
| y = 3 | never — above the whole graph | 0 |
| y = 2 | peak and endpoint | 2 |
| y = 0 | three crossings | 3 |
| y = −1.5 | only on the first slope | 1 |
If a graph has a flat piece at height k, the line y = k lies along it, so f(x) = k has infinitely many solutions — every x under that piece.
Two reads in a row, and reading a whole interval.
Same graph: corners at (−4, −2), (−2, 2), (1, −1) and (4, 2). Find f(f(−4)).
f(−4) = −2
inside first: up from x = −4
f(−2) = 2
the output becomes the next input — back on the x-axis
Stopping halfway
−2 is only f(−4). The answer after one read is always among the choices; f(f(…)) needs the second trip, starting again from the x-axis.
For which x is f(x) > 0?
zeros at x = −3, 0, 2
the crossings split the graph into pieces
above the axis: between −3 and 0, and right of 2
f(x) > 0 means the curve is ABOVE the x-axis
−3 < x < 0 or 2 < x ≤ 4
the graph ends at x = 4, so that end is included
For f(x) ≥ 0, add the zeros themselves: −3 ≤ x ≤ 0 or 2 ≤ x ≤ 4. The answer is a set of x-values — read off the x-axis, not the heights. It holds for 7 integers: −3, −2, −1, 0, 2, 3 and 4.
Against any other height k, compare with the line y = k instead of the x-axis: f(x) ≥ 2 means on or above y = 2, which here happens only at x = −2 and x = 4, where the curve touches that line.
On the exam
Graph-of-f questions mix a value f(a), a count of solutions of f(x) = k, a composite f(f(a)), and "for which x is f(x) > 0". Ask each time whether the number is an input (start on x) or an output (start on y); for inequalities, mark the zeros first and read which pieces lie above or below.
Check: the middle piece is y = −x, and it agrees.
middle piece
f(0) = −0 = 0
crosses at the origin, as drawn ✓
right piece
f(x) = x − 2 → f(3) = 1
matches the dashed read ✓
left piece
f(x) = 2x + 6 → f(−3) = 0
the first zero ✓
Now you try
Practice problem
Count the solutions
How many solutions does f(x) = −1 have?
Draw the horizontal line y = −1. It crosses the first slope at x = −3.5 (where 2x + 6 = −1) and touches the valley at (1, −1) — two meetings, so 2 solutions. A skips the touch at the valley, but a touch is still a meeting. B counts the valley twice because two pieces meet there — it is one point, so it gives one x.