Intercept

Another important property of a line (or any curve on a coordinate plane) are its x- and y-intercepts: the points where the line intersects coordinate axes. Watch this lecture series and complete the interactive exercises.

Intercepts from a table

Answers

An $x$ -intercept is a point on the line that is on the $x$ -axis, which is a point where the $y$ -value is $0$ .

For points on a line, a constant change in the $x$ -value brings a constant change in the $y$ -value. Let's use this fact to find the point where the $y$ -value is $0$ .

The table shows that for each increase of $19$ in $x$ , there's a decrease of $11$ in $y$ .

$x$	$y$
$33$	$-22$
$\stackrel{+19}{\longrightarrow} 52$	$-33 \stackrel{\longleftarrow}{-11}$
$\stackrel{+19}{\longrightarrow} 71$	$-44 \stackrel{\longleftarrow}{-11}$

Let's start at $(33,-22)$ and extend the table backwards to get to a $y$ -value of $0$ :

$x$	$y$
$33$	$-22$
$\stackrel{-19}{\longrightarrow} 14$	$-11 \stackrel{-(-11)}{\longleftarrow}$
$\stackrel{-19}{\longrightarrow} -5$	$0 \stackrel{-(-11)}{\longleftarrow}$

In conclusion, the line's $x$ -intercept is $(-5,0)$ .

To verify, here is the graph of the line. You can see it passes through all the points we've seen, including the $x$ -intercept at $(-5, 0)$ .

A $y$ -intercept is a point on the line that is on the $y$ -axis, which is a point where the $x$ -value is $0$ .

For points on a line, a constant change in the $x$ -value brings a constant change in the $y$ -value. Let's use this fact to find the point where the $x$ -value is $0$ .

The table shows that for each increase of $7$ in $x$ , there's an increase of $14$ in $y$ .

$x$	$y$
$-28$	$-54$
$\stackrel{+7}{\longrightarrow} -21$	$-40 \stackrel{\longleftarrow}{+14}$
$\stackrel{+7}{\longrightarrow} -14$	$-26 \stackrel{\longleftarrow}{+14}$

Let's start at $(-14,-26)$ and extend the table to get to an $x$ -value of $0$ :

$x$	$y$
$-14$	$-26$
$\stackrel{+7}{\longrightarrow} -7$	$-12 \stackrel{\longleftarrow}{+14}$
$\stackrel{+7}{\longrightarrow} 0$	$2 \stackrel{\longleftarrow}{+14}$

In conclusion, the line's $y$ -intercept is $(0,2)$ .

To verify, here is the graph of the line. You can see it passes through all the points we've seen, including the $y$ -intercept at $(0,2)$ .

An $x$ -intercept is a point on the line that is on the $x$ -axis, which is a point where the $y$ -value is $0$ .

For points on a line, a constant change in the $x$ -value brings a constant change in the $y$ -value. Let's use this fact to find the point where the $y$ -value is $0$ .

The table shows that for each increase of $15$ in $x$ , there's a decrease of $10$ in $y$ .

$x$	$y$
$-38$	$40$
$\stackrel{+15}{\longrightarrow} -23$	$30 \stackrel{\longleftarrow}{-10}$
$\stackrel{+15}{\longrightarrow} -8$	$20 \stackrel{\longleftarrow}{-10}$

Let's start at $(-8,20)$ and extend the table to get to a $y$ -value of $0$ :

$x$	$y$
$-8$	$20$
$\stackrel{+15}{\longrightarrow} -7$	$10 \stackrel{\longleftarrow}{-10}$
$\stackrel{+15}{\longrightarrow} -22$	$0 \stackrel{\longleftarrow}{-10}$

In conclusion, the line's $x$ -intercept is $(22,0)$ .

To verify, here is the graph of the line. You can see it passes through all the points we've seen, including the $x$ -intercept at $(22, 0)$ .

A $y$ -intercept is a point on the line that is on the $y$ -axis, which is a point where the $x$ -value is $0$ .

For points on a line, a constant change in the $x$ -value brings a constant change in the $y$ -value. Let's use this fact to find the point where the $x$ -value is $0$ .

The table shows that for each increase of $16$ in $x$ , there's an increase of $5$ in $y$ .

$x$	$y$
$32$	$40$
$\stackrel{+16}{\longrightarrow} 48$	$17 \stackrel{\longleftarrow}{-5}$
$\stackrel{+16}{\longrightarrow} 64$	$12 \stackrel{\longleftarrow}{-5}$

Let's start at $(32,22)$ and extend the table backwards to get to an $x$ -value of $0$ :

$x$	$y$
$32$	$22$
$\stackrel{-16}{\longrightarrow} 16$	$27 \stackrel{-(-5)}{\longleftarrow}$
$\stackrel{-16}{\longrightarrow} 0$	$32 \stackrel{-(-5)}{\longleftarrow}$

In conclusion, the line's $y$ -intercept is $(0,32)$ .