A zero is where the output is nothing — and that is exactly where the graph crosses the x-axis.
Find the zeros
What are the zeros of f(x) = x² − 7x + 12?
Set f(x) = 0 and factor: x² − 7x + 12 = (x − 3)(x − 4) = 0, so x = 3 or 4. A keeps the signs from inside the factors — but (x − 3) = 0 gives x = +3, not −3. That sign flip is the commonest error here. Check: f(3) = 9 − 21 + 12 = 0 ✓
| Given as | Domain | Range |
|---|---|---|
| Ordered pairs | list the x's | list the y's |
| A table | the top row | the bottom row |
| A graph | how far it goes left–right | how far up–down |
| An equation | usually all reals, minus what breaks it | from the shape |
For an equation, ask what x would break it. A denominator of x − 2 excludes 2; a square root of x − 5 needs x ≥ 5. If nothing breaks, the domain is every real number.
For a quadratic the range always starts or stops at the vertex, and which one depends on whether the parabola opens up or down.
Set f(x) = 0 and solve.
f(x) = 3x − 12
a line
3x − 12 = 0
set the output to zero
x = 4
the zero
f(x) = x² − 7x + 12
a parabola
(x − 3)(x − 4) = 0
factor
x = 3 or x = 4
two zeros
| Word | Means |
|---|---|
| Zero of the function | an x with f(x) = 0 |
| Root of the equation | a solution of f(x) = 0 |
| x-intercept | where the graph crosses the x-axis |
It is f(0) — set x to zero, not y. Here f(0) = 12, the constant term. One is found by setting the output to zero, the other the input.
"No real zeros" is a proper answer. A parabola whose lowest point is above the x-axis simply never touches it.
Factor the bottom fully: every factor that can be zero removes one value from the domain.
Find the domain of f(x) = x − 3x² − x − 6.
x² − x − 6 = (x − 3)(x + 2)
factor the bottom
x − 3 = 0 or x + 2 = 0
where the bottom is zero
all real numbers except 3 and −2
check: at x = 3 the bottom is 9 − 3 − 6 = 0 ✓
Cancelling the (x − 3) on top and bottom leaves 1x + 2, and a student then excludes only −2. But at x = 3 the original fraction is 00, which is still undefined. Read the domain from the bottom before you cancel anything.
Backwards: a parabola crosses the x-axis at 4 and −1. Write a function for it.
x = 4 means (x − 4) = 0
each zero gives a factor
x = −1 means (x + 1) = 0
the sign flips
f(x) = (x − 4)(x + 1) = x² − 3x − 4
check: f(4) = 16 − 12 − 4 = 0 ✓
Writing (x + 4)(x − 1) is the sign slip again: that function is zero at −4 and 1.
Three cases the basic rules handle, once you see them.
The domain of f(x) = √6 − 2x
6 − 2x ≥ 0
the inside may not be negative
−2x ≥ −6
subtract 6
x ≤ 3
dividing by −2 flips the sign
With −x inside the root the domain points left, not right. Check: x = 3 gives √0 = 0, allowed.
The range of f(x) = (x − 2)² − 5
(x − 2)² ≥ 0
a square is never negative
(x − 2)² − 5 ≥ −5
smallest when x = 2
range: y ≥ −5
opens up, so upward from −5
Sliding the parabola sideways changes where the vertex is, not how high it is. Only the −5 matters.
The zeros of f(x) = 2x² − x − 3
(2x − 3)(x + 1) = 0
factor
2x − 3 = 0 gives x = 32
solve, do not just flip the sign
x = 32 or x = −1
check: 2(94) − 32 − 3 = 0 ✓
When x has a coefficient in a factor, the zero is a fraction, not the bare number in the bracket.
Expect a domain with a fraction or a square root, a range read from a parabola's vertex, or zeros from a factorable quadratic — and sometimes the zeros given and the function asked. The fastest route is to name what breaks the rule (a zero bottom, a negative under the root) and to factor for zeros. The 10-second check: put each zero back in and get 0, and put an excluded value in and get a zero bottom.
Find the range
What is the range of f(x) = x² − 6?
x² is never negative, so its smallest value is 0, making the smallest value of x² − 6 equal to −6. The parabola opens upward, so everything from −6 up is reached: y ≥ −6. A is the domain, not the range. C points the inequality the wrong way — the vertex is the minimum, so values go up from it.
Hint: find the lowest point first. Since x² ≥ 0 always, the whole expression bottoms out when x = 0.
