How Slope and Y-Intercept Change a Line | Grade 8 Math | LevelKite

You already know how to graph a line from its equation. Now you will predict how the graph changes when one number in the equation changes.

Every non-vertical line has an equation of the form $y = mx + b$ , where $x$ can be any number. Such an equation is called linear because its graph is a straight line. Two numbers control the graph:

The slope $m$ is the change in $y$ divided by the change in $x$ between any two different points on the line. It tells you how many units $y$ changes each time $x$ increases by 1 unit.
The y-intercept $b$ is the $y$ -value where the line crosses the $y$ -axis. Putting $x = 0$ into the equation gives $y = m \cdot 0 + b = b$ , so the line passes through $(0, b)$ .

A vertical line such as $x = 2$ is the exception. All its points share one $x$ -value, so the change in $x$ between any two of them is 0. You cannot divide by 0, so its slope is undefined. It also gives many $y$ -values for one $x$ , so it is not a function $y = f(x)$ , which needs exactly one $y$ for each $x$ . It cannot be written as $y = mx + b$ .

Change only the y-intercept. Start with Line A, $y = 2x$ , where $b = 0$ . Keep the slope 2 and change the y-intercept to 3. The new equation is $y = 2x + 3$ . Call its graph Line B.

For any $x$ , the new equation takes the old $y$ -value, $2x$ , and adds 3. So each point on Line A has a partner on Line B with the same $x$ and a $y$ -value exactly 3 units greater. The table checks this for four values of $x$ .

Same slope, different y-intercepts

x	Line A: y = 2x	Line B: y = 2x + 3	Line B minus Line A
$-1$	$2(-1) = -2$	$2(-1) + 3 = 1$	$1 - (-2) = 3$
$0$	$2(0) = 0$	$2(0) + 3 = 3$	$3 - 0 = 3$
$1$	$2(1) = 2$	$2(1) + 3 = 5$	$5 - 2 = 3$
$2$	$2(2) = 4$	$2(2) + 3 = 7$	$7 - 4 = 3$

Line A is

y = 2x

and Line B is

y = 2x + 3

. At

x = 0

and at

x = 2

, the point on Line B is 3 units above the point on Line A.

The figure graphs both equations. The horizontal axis shows $x$ , the vertical axis shows $y$ , and each grid step is 1 unit. The marked points come from the table: $(0, 0)$ and $(2, 4)$ lie on Line A, and $(0, 3)$ and $(2, 7)$ lie on Line B. Each marked point on Line B is 3 units straight above the point on Line A with the same $x$ . Raising $b$ by 3 moved the whole line up 3 units.

Two facts follow:

The slope is unchanged. From $x = 0$ to $x = 2$ , Line A rises $4 - 0 = 4$ units and Line B rises $7 - 3 = 4$ units. Adding 3 to both $y$ -values leaves their difference the same, so both slopes are $\frac{4}{2} = 2$ .
The lines never meet. At every $x$ the two $y$ -values differ by 3, never by 0. Lines in a plane that never meet are called parallel.

Change only the slope. Now keep $b = 4$ and change $m$ . Putting $x = 0$ into $y = mx + 4$ gives $y = m \cdot 0 + 4 = 4$ for every $m$ . So the point $(0, 4)$ stays on the line, and the line tilts around it.

The sign of $m$ sets the direction. If $m > 0$ , $y$ increases as $x$ increases, so the line rises from left to right. If $m < 0$ , $y$ decreases as $x$ increases, so the line falls from left to right. If $m = 0$ , $y = 4$ for every $x$ , so the line is horizontal.
The size of $m$ without its sign, written $|m|$ , sets the steepness: the number of units $y$ changes, up or down, for each 1 unit increase in $x$ . A larger $|m|$ means a steeper line. However large $|m|$ is, the line is never vertical: a vertical line has undefined slope.

In the figure, the horizontal axis shows $x$ , the vertical axis shows $y$ , and each grid step is 1 unit. Line C is $y = x + 4$ , Line D is $y = 3x + 4$ , and Line E is $y = -3x + 4$ . All three pass through $(0, 4)$ . Move 1 unit right from there: Line C rises 1 unit to $(1, 5)$ , Line D rises 3 units to $(1, 7)$ , and Line E falls 3 units to $(1, 1)$ .

Line C is

y = x + 4

, Line D is

y = 3x + 4

and Line E is

y = -3x + 4

. All three pass through

(0, 4)

. The points at

x = 1

show how far each line has risen or fallen.

Key concept

What the slope and the y-intercept each control

For any non-vertical line $y = mx + b$ :

Change the y-intercept, keep the slope: every point moves straight up if $b$ increases, or straight down if $b$ decreases, by the size of the change. The new line is parallel to the old one and crosses the $y$ -axis at the new $(0, b)$ .
Change the slope, keep the y-intercept: the line tilts around $(0, b)$ . The sign of $m$ sets whether the line rises or falls, and $|m|$ sets how steep it is.

The relationship is proportional only when $b = 0$ , because only then does the line pass through the origin. No value of $m$ or $b$ gives a vertical line: its slope is undefined and it is not a function $y = f(x)$ . Judge these from the equations or exact points, not from how a picture looks.

y = mx + b

Worked example

Changing only the y-intercept

A line has the equation $y = 3x + 2$ . Its y-intercept is changed from 2 to $-4$ , so the new equation is $y = 3x - 4$ . Explain how the new graph differs from the original graph.

Read the slope and the y-intercept from each equation. Both equations have the form y = mx + b, so both lines are non-vertical and you can compare them part by part.
$\text{Original: } m = 3,\ b = 2 \qquad \text{New: } m = 3,\ b = -4$
Compare the slopes. Both are 3, so on both lines y rises 3 units for each 1 unit increase in x. The direction and the steepness stay the same.
Subtract the original y-value from the new y-value at the same x. The 3x terms cancel, so the result is the same for every x. It equals the change in the y-intercept.
$(3x - 4) - (3x + 2) = 3x - 4 - 3x - 2 = -6$
The result is negative, so each point on the new line is 6 units below the point on the original line with the same x. Check with exact points at x = 1: the original line gives y = 5, and the new line gives y = -1, which is 6 less than 5.
$3(1) + 2 = 5 \qquad 3(1) - 4 = -1 \qquad 5 - 6 = -1$

Answer

The new graph is the original line moved straight down 6 units. The slope is still 3, so the two lines are parallel, and the new line crosses the $y$ -axis at $(0, -4)$ instead of $(0, 2)$ .

Worked example

Changing only the slope

A line has the equation $y = \frac{1}{2}x + 3$ . Its slope is changed from $\frac{1}{2}$ to $-2$ , so the new equation is $y = -2x + 3$ . Explain how the new graph differs from the original graph.

Read the slope and the y-intercept from each equation. Both equations have the form y = mx + b, so both lines are non-vertical.
$\text{Original: } m = \tfrac{1}{2},\ b = 3 \qquad \text{New: } m = -2,\ b = 3$
Compare the y-intercepts. Both are 3, so both lines pass through the point (0, 3). Changing the slope cannot move this point, because at x = 0 the slope is multiplied by 0.
$\tfrac{1}{2}(0) + 3 = 3 \qquad -2(0) + 3 = 3$
Compare the directions using the signs of the slopes. The original slope, one half, is positive, so the original line rises from left to right. The new slope, -2, is negative, so the new line falls from left to right.
$\tfrac{1}{2} > 0 \qquad -2 < 0$
Compare the steepness using the sizes of the slopes without their signs. On the original line, y changes one half of a unit for each 1 unit increase in x. On the new line, y changes 2 units. Because 2 is greater than one half, the new line is steeper.
$\left|\tfrac{1}{2}\right| = \tfrac{1}{2} \qquad |-2| = 2 \qquad 2 > \tfrac{1}{2}$
Check with exact points at x = 2. The original line reaches (2, 4), which is 1 unit above (0, 3). The new line reaches (2, -1), which is 4 units below (0, 3).
$\tfrac{1}{2}(2) + 3 = 4 \qquad -2(2) + 3 = -1$

Answer

The new line still passes through $(0, 3)$ , but it tilts the other way: it falls from left to right instead of rising, and it is steeper because $2 > \frac{1}{2}$ .

Worked example

Two lines in standard form

Two lines are written in standard form, $Ax + By = C$ . The first line is $3x + y = 5$ and the second line is $x + y = -2$ . Without graphing, explain how the graph of the second line differs from the graph of the first.

Solve each equation for y so that you can read the slope and the y-intercept. Subtract 3x from both sides of the first equation and x from both sides of the second. Subtracting the same expression from both sides keeps an equation true for exactly the same points, so neither line changes. Both results have the form y = mx + b, so both lines are non-vertical.
$y = 5 - 3x = -3x + 5 \qquad y = -2 - x = -x - 2$
Compare the y-intercepts: 5 for the first line and -2 for the second. The crossing point on the y-axis moves from (0, 5) down to (0, -2), a distance of 7 units. The slope changes too, so the rest of the line does not simply slide down 7 units.
$5 - (-2) = 5 + 2 = 7$
Compare the slopes. The first line has slope -3. The second has slope -1, because -x means -1 times x. Both slopes are negative, so both lines fall from left to right. For steepness, compare the sizes without signs: 1 is less than 3, so the second line is less steep, even though -1 is greater than -3.
$|-3| = 3 \qquad |-1| = 1 \qquad 1 < 3$
Check with exact points at x = 1. The first line reaches (1, 2), a fall of 3 units from (0, 5). The second line reaches (1, -3), a fall of 1 unit from (0, -2). Both points also make the original equations true: 3(1) + 2 = 5 and 1 + (-3) = -2.
$-3(1) + 5 = 2 \qquad -(1) - 2 = -3$

Answer

The second line crosses the $y$ -axis at $(0, -2)$ , which is 7 units below the first line's crossing point $(0, 5)$ . Both lines fall from left to right, but the second line is less steep because $1 < 3$ .

Takeaway. For a non-vertical line $y = mx + b$ , each number has its own job. Change only $b$ , and the whole line slides straight up or down and stays parallel to where it was. Change only $m$ , and the line tilts around $(0, b)$ : the sign of $m$ decides whether it rises or falls, and $|m|$ decides how steep it is. The line shows a proportional relationship only when $b = 0$ . When you compare two lines, read $m$ and $b$ from the equations and confirm with exact points, not with how a picture looks.