How to Find X-Intercepts
An x-intercept is a point where a graph meets the x-axis. Every point on that axis has y = 0, so finding them always comes down to one move: set y = 0 and solve for x. What changes from one kind of function to the next is how you do the solving.
The rule
Set y = 0 and solve for x. Each solution a gives an x-intercept at the point (a, 0). For a function written y = f(x), that means solving f(x) = 0, which is why x-intercepts are also called the zeros or roots of the function.
Give the answer as a point if the question asks for intercepts — (3, 0) rather than 3 — and as a value if it asks for zeros or roots.
Lines
For y = 2x + 6, setting y = 0 gives 2x + 6 = 0,
so x = −3, and the intercept is (−3, 0).
For a line in standard form, 2x + 3y = 6, put y = 0 directly:
2x = 6, so (3, 0). This is the quickest way to
graph a line in that form — find both intercepts by covering up each
variable in turn, then join them.
Every line meets the x-axis exactly once, except horizontal lines: y = 5 never does, and y = 0 is the x-axis itself.
Quadratics
Three ways, in the order worth trying them.
Factor
x² − 5x + 6 = (x − 2)(x − 3). A product is zero
only when one of its factors is, so the intercepts are
(2, 0) and (3, 0).
Use the quadratic formula
When it will not factor over the whole numbers, use
x = (−b ± √(b² − 4ac)) / 2a. For
x² − 2x − 1, a = 1, b = −2, c = −1:
x = (2 ± √8) / 2 = 1 ± √2
So the intercepts are at 1 − √2 and 1 + √2, about −0.414 and 2.414. Leave them as surds unless decimals are asked for.
Check the discriminant first
The part under the square root, b² − 4ac, tells you how many intercepts there are before you find them.
- Positive: two. x² − 2x − 1 has 8.
- Zero: one, where the parabola just touches the axis. x² − 4x + 4 has 0, and touches at (2, 0).
- Negative: none. x² + 2x + 5 has −16, and the parabola sits entirely above the axis.
Higher polynomials
Factor as far as you can. Take out any common factor first, then look for a pattern or a root you can test.
-
x³ − 2x = x(x² − 2), so x = 0 or x² = 2: the intercepts are at −√2, 0 and √2. -
x⁴ − 5x² + 6looks hard but is a quadratic in x². Writing u = x² gives u² − 5u + 6 = (u − 2)(u − 3), so x² = 2 or x² = 3, and the intercepts are at ±√2 and ±√3.
That second one is worth a note: it has no whole-number or fractional roots at all, so a method that only hunts for those finds nothing, and many online solvers fall back to decimals. It factors anyway — into two quadratics — and each gives exact roots. The intercept calculator factors completely first for exactly this reason.
For a polynomial that will not factor neatly, try whole numbers that divide the constant term — by the rational root theorem, any whole-number root must be one of them — and divide out each one you find.
Crossing or touching
How the graph meets the axis depends on how many times the factor appears.
In (x − 2)²(x + 1):
-
(x + 1)appears once, so the graph crosses at x = −1. -
(x − 2)appears twice, so the graph comes down to the axis at x = 2 and turns back without crossing.
The rule: odd number of times, crosses; even, touches. It matters for sketching, and for any question that asks where the function changes sign.
Fractions: watch for holes
A fraction is zero when its top is zero and its bottom is not. So for a rational function, find the zeros of the numerator, then check each one against the denominator.
In (x² − 1)/(x − 1) the numerator is zero at
x = 1 and x = −1. But at x = 1 the denominator is zero as well: the
factor (x − 1) cancels, and the graph is the line y = x + 1 with a
hole at (1, 2). The only x-intercept is (−1, 0).
The sharpest version is (x − 1)²/(x − 1). Cancel
and you get x − 1, which is zero at x = 1 — so it looks like an
intercept at (1, 0). But the original function is undefined there. The
graph is the line y = x − 1 with the point (1, 0) missing: the hole is
sitting on the axis, exactly where you would draw the intercept, and the
function has no x-intercept at all.
Exponentials, logs and trig
-
eˣ − 2 = 0gives eˣ = 2, so x = ln 2 ≈ 0.693. -
ln x = 0gives x = 1, since ln 1 = 0. Every log graph of the form y = logb(x) crosses at (1, 0). -
sin x = 0at every multiple of π: x = 0, ±π, ±2π and so on, infinitely many intercepts. -
eˣon its own is never zero, so it has no x-intercept.
Circles and other equations
The rule is the same for any equation in x and y: put y = 0 and solve.
For the circle x² + y² = 25, y = 0 gives x² = 25, so there are
two intercepts, (−5, 0) and
(5, 0).
For a shifted circle like (x − 1)² + (y − 2)² = 9,
y = 0 gives (x − 1)² = 5, so x = 1 ± √5.
Mistakes that cost marks
- Setting x = 0 instead of y = 0. That finds the y-intercept.
- Dividing by x to simplify, and losing the root x = 0. In x³ − 2x = 0, factor out x rather than dividing it away.
- Counting a hole as an intercept in a rational function.
- Giving decimals when exact values exist. 1 ± √2 is the answer; −0.414 and 2.414 are approximations of it.
- Reading intercepts off a graph and reporting the pixel. A graph shows roughly where an intercept is; algebra tells you exactly.
Questions
What is another name for the x-intercepts?
For a function y = f(x), the x-coordinates of its x-intercepts are its zeros, also called its roots: the values of x that make f(x) = 0. They are also the solutions of the equation f(x) = 0. The x-intercepts themselves are points, written with a y-coordinate of 0, such as (3, 0).
How many x-intercepts can a function have?
A polynomial of degree n has at most n, because it has at most n roots. A line that is not horizontal has exactly one. Other functions can have infinitely many: sin x meets the x-axis at every multiple of π. And many functions have none at all.
Can a function have no x-intercepts?
Yes. x² + 1 is never less than 1, eˣ is always positive, and the horizontal line y = 5 never reaches the axis. For a quadratic, a negative discriminant b² − 4ac is the test: it means the parabola stays entirely above or entirely below the x-axis.
What is the difference between an x-intercept that crosses and one that touches?
At a crossing the graph passes from one side of the axis to the other; at a touch it comes down to the axis and turns back. A root repeated an odd number of times is crossed, and one repeated an even number of times is touched. In (x − 2)²(x + 1) the graph crosses at x = −1 and touches at x = 2.
Why is my answer not an x-intercept when the numerator is zero?
Because the denominator is probably zero there too. If a factor cancels between the top and bottom of a fraction, the graph has a hole at that x-value, not a point, and a missing point cannot be an intercept. Always check each zero of the numerator against the denominator before calling it an intercept.