The intercepts of graphs

Consider the graph of a function $f$ . Each point on the $x$ -axis where the graph touches or crosses is called an x-intercept ( $x$ -Achsenabschnitt), and likewise, each point on the $y$ -axis where the graph touches or crosses is called a y-intercept ( $y$ -Achsenabschnitt).

For example, the graph below has $x$ -intercepts at $x=-2$ , $x=1$ , and $x=4$ , and a $y$ -intercept at $y=2$ .

Exercise 1

What is the lowest and highest number of possible $x$ -intercepts and $y$ -intercepts?

Solution

You can have any number of $x$ -intercepts, including none. The number of $y$ -intercepts, however, is either $0$ or $1$ . Why is it not possible to have more than one $y$ -intercept? Because it's a function! If there is more than one $y$ -intercept, the input $0$ will have more than $1$ output. But then the graph does not represent a function any more.

Now, consider a function whose function equation is known, e.g.

f(x)=x^2-2

How can we find the $x$ -and $y$ -intercepts of its graph? The $y$ -intercept is on the $y$ -axis and therefore is the output to the input $x=0$ . Thus, the $y$ -intercept is

\boxed{y-\text{intercept} = f(0)}

Each $x$ -intercept is an input value such that the output is $0$ :

\boxed{x\text{-intercept: all inputs $x$ with } f(x)=0}

Because $x$ -intercepts indicate the inputs where the output of the function is zero, we also call them the zeros of $f$ .

Example 1

Find the $x$ - and $y$ -intercept of $f(x)=x^2-2$ .

The $y$ -intercept is $y=f(0)=0^2-2=\underline{-2}$
The $x$ - intercepts are found by solving the equation $f(x)=0$ , that is

x^2-2=0

It follows that $x=\pm \sqrt{2}$ , that is, we have the two $x$ -intercepts $x_1=\underline{1.414...}$ and $x_2=\underline{-1.414...}$

Indeed, a plot of the graph shows exactly this (uncollapse)

Solution

Exercise 2

Q1

Consider the function $f(x)=x(x-2)$ .

Draw the graph of $f$ .
Based on the graph, estimate the $x$ -intercepts and the $y$ -intercept.
Calculate the $x$ -intercepts and the $y$ -intercept.

Q2

Calculate the $x$ -intercept(s) and the $y$ -intercept of the following functions

$f(x)=3x+1$
$g(x)=2x^2-9$
$h(x)=x^2+1$
$i(x)=2x^2-3x$
$j(x)=4x(x+1)(2x-3)$
$k(x)=\frac{2}{x^2}+1$

Q3

Consider the functions $f(x)=x^2$ and $g(x)=0.5x+1$ .

Draw the graphs of the two functions into the same coordinate system.
Calculate their $x$ -intercepts and $y$ -intercept.
The two graphs intersect. Estimate the points of intersection ( $x$ - and $y$ -coordinates).

Q4

Find the zeros of the following functions:

$f(x)=(x-1)(x-2)(x-3)(x-4)$
$g(x)=\sqrt{x^2-16}$
$h(x)=x^4+x^3$

Solution

A1

The graph is shown below
Graph $\rightarrow x_1\approx\underline{0}$ and $x_2\approx\underline{2}$
$x$ -intercepts: find $x$ with $f(x)=x(x-2)=0\rightarrow x_1=\underline{0}$ and $x_2=\underline{2}$ . $y$ -intercept: $y=f(0)=0\cdot (0-2)=\underline{0}$ .

A2

$x$ -intercepts: $f(x)=3x+1=0\rightarrow x=\underline{-\frac{1}{3}}$ . $y$ -intercept: $y=f(0)=3\cdot 0+1=\underline{1}$
$x$ -intercepts: $g(x)=2x^2-9=0\rightarrow x^2=4.5 \rightarrow x=\underline{\pm\sqrt{4.5}}$ . $y$ -intercept: $y=g(0)=2\cdot 0^2-9=\underline{-9}$
$x$ -intercepts: $h(x)=x^2+1=0\rightarrow x^2=-1 \rightarrow x=\sqrt{-1} \rightarrow$ no solution, so no $x$ -intercept. $y$ -intercept: $y=h(0)=0^2+1=\underline{1}$
$x$ -intercepts: $i(x)=2x^2-3x=0\rightarrow x_1=\underline{0}$ and $x_2=\underline{1.5}$ . $y$ -intercept: $y=i(0)=\underline{0}$
$x$ -intercepts: $j(x)=4x(x+1)(2x-3)=0\rightarrow x_1=\underline{0}$ and $x_2=\underline{-1}$ and $x_3=\underline{1.5}$ (each factor has to be zero). $y$ -intercept: $y=j(0)=\underline{0}$
$x$ -intercepts: $k(x)=\frac{2}{x^2}+1=0\rightarrow 2=-x^2 \rightarrow x^2=-2 \rightarrow x=\sqrt{-2} \rightarrow$ no solution, so no $x$ -intercept. $y$ -intercept: $y=k(0)=\frac{2}{0}+1$ . As division by $0$ is infinity, so no real number, there is also no $y$ -intercept.

A3

The graph is shown below.
$x$ -intercepts of $f$ : $f(x)=x^2=0\rightarrow x=\underline{0}$ . $y$ -intercept of $f$ : $y=f(0)=\underline{0}$ . $x$ -intercepts of $g$ : $g(x)=0.5x+1=0\rightarrow x=\underline{-2}$ . $y$ -intercept of $g$ : $y=g(0)=\underline{1}$
Estimates based on figure: $A(-0.9|0.8)$ , $B(1.1|1.8)$ .

A4

Find $x$ with $f(x)=(x-1)(x-2)(x-3)(x-4)=0$ . So each factor has to be zero. It follows $x_1=\underline{1}, x_2=\underline{2}, x_3=\underline{3}, x_4=\underline{4}$ .
Find $x$ with $\sqrt{x^2-16}=0$ . Square on both sides, so we get $x^2-16=0^2=0$ and thus $x^2=16$ . Taking the root, we get the two zeros $x_1=\underline{4}, x_2=\underline{-4}$ .
Find $x$ with $x^4+x^3=0$ . Let's factor out an $x^3$ , thus we get $x^3(x+1)=0$ and because each factor has to be zero, we get $x_1=\underline{0}, x_2=\underline{-1}$ .