Algebra 1 · F.IF.B.4 · F.IF.C.7A

Intercepts and zeros of a function

Where does a graph meet the two axes, and what do those meeting points say about the function?

You need

Unlocks

Increasing, decreasing, constant, positive and negative intervalsAlgebra 1 Domain and zeros of a rational functionAlgebra 2

Where you are 10 in Features of a graph

Functions13

What a function is5

Graphs of functions5

Features of a graph10

End behavior Even and odd functions

Intercepts and zeros of a function

Symmetry about the vertical axis and the origin What an asymptote is ▾ Show all 10 (5 more)

Why this exists

Roots, zeros and x-intercepts gave three names to one set of numbers (root, zero, x-intercept) and found them for quadratics in factored form. Nothing about those names belongs to quadratics. Every function has a graph, and the two places a graph can meet the axes carry meaning in almost any setting. Where the graph meets the horizontal axis the output is 0: a lift is level with the ground floor, a phone has used up its data. Where it meets the vertical axis the input is 0: the clock has just started, so the output there is the starting amount. This lesson finds both kinds of point for any function, from a graph, from a table and from a rule.

Definition

Let f be a function.

An x-intercept of the graph of f is a point where the graph meets the horizontal axis. Every point on that axis has second coordinate 0, so an x-intercept is a point (k, 0) with f(k) = 0.
A zero of f is an input k with f(k) = 0. The zero is the number k; the x-intercept is the point (k, 0).
The y-intercept of the graph of f is the point where the graph meets the vertical axis. Every point on that axis has first coordinate 0, so the y-intercept is (0, f(0)). It exists only when 0 is in the domain of f.

Finding f(0) is evaluating the function at one number. Finding the zeros is solving the equation f(x) = 0, which may have no solution, one, or many. When the variables have other names, such as t and h, the same points are called the horizontal intercept and the vertical intercept.

Intuition

The height of the graph above the axis is the output. A zero is an input where that height is exactly 0, so the graph sits on the horizontal axis there. The vertical axis is the vertical line x = 0, and the graph of a function meets any vertical line at most once (the Testing graphs and equations for functions), because one input has only one output. So there is at most one y-intercept. The horizontal axis is not a vertical line, and nothing stops a graph meeting it several times, or never.

Figure 1Three x-intercepts but only one y-intercept.

A graph may also come down to the axis, touch it and turn back without crossing. That point is still an x-intercept; which zeros cross and which only touch is the subject of Zeros and multiplicity.

A table shows only some inputs. A 0 in the output column gives a zero exactly. Between two rows whose outputs have opposite signs, the graph goes from one side of the axis to the other. If it is drawn without lifting the pen (an unbroken graph) it must meet the axis somewhere in between. A table alone cannot say whether the graph is unbroken, where between the two rows the zero sits, or whether another zero hides between two rows of the same sign.

Results

At most one y-intercept. If 0 is in the domain of f, the graph has exactly one y-intercept, (0, f(0)). If not, it has none.

Any number of x-intercepts. A function may have no zeros, one, several, or infinitely many. f(x) = 5 has none; f(x) = 0 has every real number as a zero.

Lines. The linear function f(x) = mx + b has y-intercept (0, b), and when m \neq 0 it has exactly one zero, \displaystyle x = -\frac{b}{m}.

A sign change. If the graph of f is unbroken from x = a to x = b, and f(a) and f(b) have opposite signs, then f has at least one zero strictly between a and b.

Why it is true

One y-intercept. The points of the vertical axis are exactly the points with first coordinate 0. A function gives one output for each input in its domain, so the input 0 gives one output, f(0), and the graph has exactly one point with first coordinate 0. If 0 is not in the domain, the graph has no such point.

Lines. f(0) = m \cdot 0 + b = b. For the zero, mx + b = 0 gives mx = -b (subtract b from both sides), and dividing by m, allowed because m \neq 0, gives the single solution \displaystyle x = -\frac{b}{m}.

A sign change. Say f(a) < 0 < f(b). At x = a the graph is below the axis and at x = b it is above. An unbroken curve cannot get from one side of a line to the other without meeting the line, so somewhere between a and b its height is 0. A full proof needs a careful meaning of “unbroken” and comes later as the intermediate value theorem. Without the unbroken condition the result fails: the function equal to -1 for x < 0 and to 1 for x \ge 0 changes sign between -1 and 1 and has no zero at all, because its graph jumps over the axis.

Worked examples

Example 1

Reading intercepts off a graph

Figure 1 is the graph of \displaystyle f(x) = \frac{1}{4}(x + 3)(x - 1)(x - 4). Read its zeros and both kinds of intercept, then confirm them with the rule.

Solution

The graph meets the horizontal axis three times, above -3, 1 and 4, so the zeros are -3, 1 and 4 and the x-intercepts are (-3, 0), (1, 0) and (4, 0). It meets the vertical axis once, at height 3, so the y-intercept is (0, 3).

A picture can only suggest a number: a crossing that looks like 1 might be 1.02. The rule settles it. \displaystyle f(1) = \frac{1}{4}(4)(0)(-3) = 0, because a product with a factor 0 is 0, and the same reason gives f(-3) = 0 and f(4) = 0. Then \displaystyle f(0) = \frac{1}{4}(3)(-1)(-4) = \frac{1}{4} \cdot 12 = 3, so the y-intercept is (0, 3) exactly.

Example 2

Data running out

Delmar’s phone has D(d) = 12 - 0.4d gigabytes of data left after d days of the month, for 0 \le d \le 30. Find both intercepts and say what each means.

Solution

The vertical intercept: D(0) = 12 - 0.4 \cdot 0 = 12, so it is (0, 12). At the start of the month Delmar has 12 gigabytes.

The horizontal intercept comes from D(d) = 0:

\begin{aligned} 12 - 0.4d &= 0 \\ 12 &= 0.4d \\ d &= 30 \end{aligned}

adding 0.4d to both sides, then dividing both sides by 0.4. So it is (30, 0): the data runs out after 30 days. Check: D(30) = 12 - 12 = 0. This is the line with slope -0.4 and vertical intercept 12 from Slope-intercept form, and the Results give the zero as \displaystyle -\frac{12}{-0.4} = 30 too.

Example 3

An absolute value function

Find the intercepts of f(x) = |x - 3| - 2.

Solution

The y-intercept: f(0) = |0 - 3| - 2 = 3 - 2 = 1, so it is (0, 1).

The zeros solve |x - 3| - 2 = 0, that is |x - 3| = 2 after adding 2 to both sides. The numbers with absolute value 2 are 2 and -2, so either x - 3 = 2, giving x = 5, or x - 3 = -2, giving x = 1. Check: f(5) = |2| - 2 = 0 and f(1) = |-2| - 2 = 0. The x-intercepts are (1, 0) and (5, 0).

Figure 2The V dips below the axis, so each arm crosses it once.

The graph is the absolute value V moved 3 right and 2 down, so its lowest point is (3, -2). That point is below the axis and both arms climb from it, so each arm crosses the axis once. For |x - 3| + 2 the lowest output would be 2 and there would be no zero.

Example 4

An exponential function with no zero

Find the intercepts of y = 4 \cdot 2^x.

Solution

The y-intercept: 4 \cdot 2^0 = 4 \cdot 1 = 4, so it is (0, 4).

A zero would need 4 \cdot 2^x = 0. But 2^x is positive for every x: for x = 1, 2, 3, \ldots it is a product of 2s, 2^0 = 1, for negative x it is \displaystyle \frac{1}{2^{-x}}, a fraction with positive top and bottom, and the graph of 2^x lies above the axis everywhere. Four times a positive number is positive, never 0. So there is no zero and no x-intercept.

Figure 3Every output is positive, so the graph never meets the horizontal axis.

Moving left, each step of 1 halves the output: 4, 2, 1, 0.5, \ldots. The outputs get as close to 0 as you like and never reach it, so the axis is an asymptote. On a drawn graph the curve seems to merge with the axis at the far left, but at x = -5 the height is still \displaystyle 4 \cdot 2^{-5} = \frac{1}{8}, and at x = -10 it is \displaystyle \frac{1}{256}, thinner than the pen yet not 0. The rule settles what the picture cannot.

Example 5

Zeros from a table

A function g with an unbroken graph has these values.

x	g(x)
-3	15
-2	4
-1	-3
0	-6
1	-5
2	0
3	9

What does the table say about the intercepts? Then use the rule g(x) = 2x^2 - x - 6 to finish.

Solution

The row x = 0 gives the y-intercept, (0, -6). The row x = 2 has output 0, so 2 is a zero and (2, 0) is an x-intercept.

From x = -2 to x = -1 the output goes from 4 to -3, a change of sign. The graph is unbroken, so at least one zero lies strictly between -2 and -1. The table does not say where, and it cannot rule out a zero between two rows of the same sign, where the graph might dip across the axis and back.

The rule settles both. Factoring, 2x^2 - x - 6 = (2x + 3)(x - 2), since (2x + 3)(x - 2) = 2x^2 - 4x + 3x - 6. A product is 0 only when a factor is, as in Roots, zeros and x-intercepts, so the zeros are \displaystyle -\frac{3}{2} and 2. The first sits between -2 and -1 as promised, and a quadratic has at most two zeros, so nothing else hides between the rows.

Example 6

A lift that goes below the ground floor

A lift’s height above the ground floor, in meters, t seconds after it starts, is

h(t) = \begin{cases} 9 - 1.5t, & 0 \le t \le 8 \\ -3, & 8 < t < 12 \\ -3 + 1.5(t - 12), & 12 \le t \le 16 \end{cases}

A negative height means below the ground floor. Find the intercepts and say what they mean.

Solution

The vertical intercept: 0 is in the first piece’s stretch, so h(0) = 9 - 0 = 9, giving (0, 9). The lift starts 9 meters above the ground floor.

For the zeros, solve each piece and keep only a solution that lies in that piece’s own stretch of t.

First piece: 9 - 1.5t = 0 gives t = 6, and 0 \le 6 \le 8, so it counts.
Second piece: -3 = 0 is never true. The lift waits 3 meters down, in the basement.
Third piece: -3 + 1.5(t - 12) = 0 gives 1.5(t - 12) = 3, so t - 12 = 2 and t = 14, which lies in 12 \le t \le 16.

Figure 4The lift passes the ground floor twice, once on the way down and once on the way up.

The horizontal intercepts are (6, 0) and (14, 0): the lift passes the ground floor on the way down after 6 seconds and on the way up after 14 seconds. Each piece has at most one zero, yet the function has two. The pieces meet (h(8) = 9 - 12 = -3 and h(12) = -3), so the graph is unbroken, and it ends at h(16) = 3, three meters above the ground floor.

Boundaries

A y-intercept needs 0 in the domain. A function defined only for 1 \le x \le 5 has none, because its graph never reaches the line x = 0.
A zero is an input where the output is exactly 0. A graph or table can suggest one; only the rule confirms it, and a curve that seems to reach the axis may not (Example 4).
The sign change result needs an unbroken graph, and it promises at least one zero, never exactly one. Equal signs at two rows do not rule a zero out: the graph may touch the axis between them, or cross twice.
In a model, an intercept means something only inside the domain the model is used for. The rule 12 - 0.4d gives 14 at d = -5, but there is no day -5 of the month.

Where this goes next

The zeros cut the input line into stretches where the graph is above the axis and stretches where it is below; Increasing, decreasing, constant, positive and negative intervals names these the positive and negative intervals. Whether a graph crosses or only touches at a zero is Zeros and multiplicity. Solving f(x) = g(x) is finding the zeros of f(x) - g(x), which is how Solving f(x) = g(x) with graphs and tables reads meeting points. For a rational function, Domain and zeros of a rational function separates the zeros of the top, which are zeros of the function, from the zeros of the bottom, which are left out of the domain.

Exercises

Graphs and tables

Figure 5 shows y = q(x). Read the y-intercept and every x-intercept. At which x-intercept does the graph cross the axis, and at which does it only touch?

Figure 5The graph for Exercise 1. Each tick mark is one unit.

Give the intercepts of f(x) = 4 and of g(x) = 0.
A function with an unbroken graph takes the values 5, 2, -1, -2, 0, 4 at x = -2, -1, 0, 1, 2, 3. Give the y-intercept, any zero the table shows exactly, and any interval that must contain a zero.
A function with an unbroken graph takes the values 6, 2, -1, -3, 1 at x = 1, 2, 3, 4, 5. At least how many zeros lie between 1 and 5? Does the table give the y-intercept?

From a rule

In Exercises 5 to 12, find the y-intercept and every x-intercept, or explain why there is none.

f(x) = 4x - 18
h(x) = |x + 1| - 4
k(x) = |3x - 6|
j(x) = |x - 4| + 1
p(x) = 7 \cdot 3^x
r(x) = 2^x - 8
s(x) = x + 4 for x < 0, and s(x) = 6 - 2x for x \ge 0
w(x) = 2x - 1 for x < 0, and w(x) = x^2 - 4 for x \ge 0

In context

Diantha’s phone has R(d) = 15 - 0.6d gigabytes left after d days. Find both intercepts and say what each means, with units.
A light dims so that its brightness after t seconds is \displaystyle B(t) = 640 \cdot \left(\frac{1}{2}\right)^t lumens. Find the vertical intercept and its meaning. Is there a horizontal intercept? After how many seconds is the brightness 5 lumens?
A cool box warms so that its temperature after t hours is T(t) = -6 + 2t degrees Celsius, for 0 \le t \le 9. Find both intercepts and say what each means.

Explain

A circle of radius 2 centered at the origin meets the vertical axis at (0, 2) and (0, -2). Explain why it is not the graph of a function, and why no function has two y-intercepts.
All that is known about f is that f(1) = -2 and f(4) = 3. Explain why this does not show that f has a zero between 1 and 4, and why, even with an unbroken graph, it does not show exactly one.
Explain why f(x) = mx + b with m \neq 0 has exactly one zero, and say what happens when m = 0.

Answers

The y-intercept is (0, 1); the x-intercepts are (-1, 0) and (3, 0). The graph touches the axis at (-1, 0) and turns back up; it crosses at (3, 0). (The figure is \displaystyle q(x) = -\frac{1}{3}(x + 1)^2(x - 3), and q(0) = 1, q(-1) = 0, q(3) = 0.)
For f: y-intercept (0, 4), no x-intercept, since every output is 4, never 0. For g: y-intercept (0, 0), and every real number is a zero, so every point of the horizontal axis is an x-intercept.
y-intercept (0, -1). The table shows the zero 2 exactly, so (2, 0). The outputs go from 2 to -1 between x = -1 and x = 0, so a zero lies between -1 and 0.
At least two: the sign changes between 2 and 3 (from 2 to -1) and between 4 and 5 (from -3 to 1). The table has no row for x = 0, so it does not give the y-intercept, or even say whether 0 is in the domain.
f(0) = -18, so (0, -18). 4x - 18 = 0 gives \displaystyle x = \frac{9}{2}, so \displaystyle \left(\frac{9}{2}, 0\right).
h(0) = |1| - 4 = -3, so (0, -3). |x + 1| = 4 gives x + 1 = 4 or x + 1 = -4, so x = 3 or x = -5: (3, 0) and (-5, 0).
k(0) = |-6| = 6, so (0, 6). Only 0 has absolute value 0, so 3x - 6 = 0 and x = 2: one x-intercept, (2, 0), where the V’s point touches the axis.
j(0) = |-4| + 1 = 5, so (0, 5). |x - 4| + 1 = 0 would need |x - 4| = -1, impossible since an absolute value is never negative. Every output is at least 1: no x-intercept.
p(0) = 7 \cdot 1 = 7, so (0, 7). 3^x > 0 for every x, so 7 \cdot 3^x > 0: no x-intercept.
r(0) = 1 - 8 = -7, so (0, -7). 2^x = 8 = 2^3 gives x = 3, and no other input works because 2^x grows as x grows: (3, 0).
0 \ge 0, so s(0) = 6 - 0 = 6: (0, 6). x + 4 = 0 gives x = -4, which is below 0, so it counts; 6 - 2x = 0 gives x = 3, which is at least 0, so it counts. x-intercepts (-4, 0) and (3, 0).
w(0) = 0^2 - 4 = -4: (0, -4). 2x - 1 = 0 gives \displaystyle x = \frac{1}{2}, not below 0, so rejected. x^2 - 4 = 0 gives x = 2 or x = -2, and -2 is not at least 0, so rejected. One x-intercept, (2, 0).
R(0) = 15: (0, 15), so 15 gigabytes at the start. 15 - 0.6d = 0 gives d = 25: (25, 0), so the data runs out after 25 days.
B(0) = 640: (0, 640), the brightness of 640 lumens when the dimming starts. No horizontal intercept: \displaystyle \left(\frac{1}{2}\right)^t > 0 for every t, so the light never goes fully dark in this model. \displaystyle 640 \cdot \left(\frac{1}{2}\right)^t = 5 gives 2^t = 128 = 2^7, so t = 7 seconds.
T(0) = -6: (0, -6), so the box starts at 6 degrees Celsius below zero. -6 + 2t = 0 gives t = 3: (3, 0), so it reaches 0 degrees Celsius after 3 hours.
Two points on the vertical axis would mean the input 0 has two outputs, 2 and -2. A function gives one output per input, so the circle is not the graph of a function; it fails the vertical line test on the line x = 0 itself. The same reason rules out two y-intercepts for any function.
Without an unbroken graph nothing is promised: a function equal to -2 for x < 3 and 3 for x \ge 3 matches both values and has no zero. Even with an unbroken graph the promise is at least one zero, not exactly one; the graph could cross three times.
With m \neq 0, mx + b = 0 is the same as \displaystyle x = -\frac{b}{m} (subtract b, divide by m), a single number. With m = 0, f(x) = b is constant: if b \neq 0 there is no zero, and if b = 0 every real number is a zero.