The Central Idea
Consider
\[ y=3x+5. \]Two pieces of information are visible immediately.
The coefficient of \(x\) is
\[ \boxed{m=3}, \]so the output increases by \(3\) units for each increase of \(1\) unit in the input.
The constant term is
\[ \boxed{b=5}, \]so when
\[ x=0, \]we have
\[ y=5. \]Therefore, the graph has \(y\)-intercept
\[ \boxed{(0,5)}. \]Slope-intercept form packages both ideas into one equation:
\[ \boxed{y=mx+b}. \]What Is Slope-Intercept Form?
A nonvertical affine line can be written in the form
\[ \boxed{y=mx+b}, \]where:
- \(m\) is the slope;
- \(b\) is the \(y\)-intercept value.
The corresponding \(y\)-intercept point is
\[ \boxed{(0,b)}. \]Because the slope and \(y\)-intercept are exposed directly, this representation is called slope-intercept form.
A Note About “Linear” and “Affine”
In many elementary algebra courses, every equation of the form
\[ y=mx+b \]is described as linear.
Malone Math recognizes that common usage.
In the stricter structural sense used throughout this series:
\[ \boxed{y=mx} \]is linear, while
\[ \boxed{y=mx+b} \]with \(b\ne0\) is affine.
Slope-intercept form works for both.
The Meaning of \(m\)
The parameter
\[ m \]is the slope, or constant rate of change.
For any two distinct points on the line,
\[ (x_1,y_1) \]and
\[ (x_2,y_2), \]we have
\[ \boxed{ m= \frac{y_2-y_1}{x_2-x_1} }. \]Equivalently,
\[ \boxed{ \Delta y=m\,\Delta x }. \]The Meaning of \(b\)
Set
\[ x=0 \]in
\[ y=mx+b. \]Then
\[ y=m(0)+b=b. \]So the graph contains
\[ \boxed{(0,b)}. \]The parameter \(b\) therefore determines the vertical position of the line when the input is zero.
Rate and Starting Value
Slope-intercept form separates two different roles:
\[ \boxed{y=mx+b}. \]- \(m\) describes how the output changes.
- \(b\) describes where the output begins when \(x=0\).
This distinction is one of the most useful features of affine models.
Example: Read the Equation
Consider
\[ y=-4x+7. \]The slope is
\[ \boxed{-4}. \]The \(y\)-intercept is
\[ \boxed{(0,7)}. \]So as \(x\) increases by \(1\), the output decreases by \(4\), and the graph crosses the \(y\)-axis at \(7\).
Example: Fractional Slope
Consider
\[ y=\frac23x-5. \]The slope is
\[ \boxed{\frac23}. \]One convenient interpretation is:
\[ \Delta x=3, \qquad \Delta y=2. \]The \(y\)-intercept is
\[ \boxed{(0,-5)}. \]Example: Negative Fractional Slope
Consider
\[ y=-\frac54x+2. \]The slope is
\[ -\frac54. \]One convenient movement along the line is
\[ \Delta x=4, \qquad \Delta y=-5. \]The graph therefore moves \(4\) units right and \(5\) units down from any point on the line.
Its \(y\)-intercept is
\[ \boxed{(0,2)}. \]Graphing from Slope-Intercept Form
The form
\[ y=mx+b \]suggests an efficient graphing procedure.
- Plot the \(y\)-intercept \((0,b)\).
- Interpret \(m\) as rise over run.
- Use that change to locate another point.
- Draw the straight line through the points.
The equation itself supplies both the starting point and the direction of change.
Graphing Example
Graph
\[ y=2x-3. \]The \(y\)-intercept is
\[ (0,-3). \]The slope is
\[ 2=\frac21. \]Starting at \((0,-3)\), move:
- right \(1\);
- up \(2\).
This gives another point:
\[ (1,-1). \]Repeat the same change to obtain
\[ (2,1). \]These points lie on the same straight line.
Graphing with a Negative Slope
Consider
\[ y=-\frac32x+4. \]Start at the \(y\)-intercept
\[ (0,4). \]Interpret
\[ -\frac32 \]as
\[ \frac{-3}{2}. \]Move right \(2\) and down \(3\):
\[ (0,4)\longrightarrow(2,1). \]Another repetition gives
\[ (4,-2). \]Equivalent Rise-and-Run Movements
The slope
\[ -\frac32 \]can also be written
\[ \frac{3}{-2}. \]So instead of moving right \(2\) and down \(3\), we may move left \(2\) and up \(3\).
Both movements stay on the same line because both represent the same ratio.
Constructing an Equation from Slope and \(y\)-Intercept
If the slope and \(y\)-intercept are known, substitute them directly into
\[ y=mx+b. \]Suppose
\[ m=5 \]and the \(y\)-intercept is
\[ (0,-2). \]Then
\[ b=-2, \]so the equation is
\[ \boxed{y=5x-2}. \]Be Careful with the Sign of \(b\)
If the \(y\)-intercept is
\[ (0,-7), \]then
\[ b=-7. \]With slope
\[ m=3, \]the equation is
\[ \boxed{y=3x-7}, \]not
\[ y=3x+7. \]Constructing an Equation from a Table
Suppose:
| \(x\) | \(y\) |
|---|---|
| 0 | 4 |
| 2 | 10 |
| 5 | 19 |
The row with \(x=0\) gives
\[ b=4. \]Find the slope:
\[ m= \frac{10-4}{2-0} = 3. \]Therefore,
\[ \boxed{y=3x+4}. \]When the Table Does Not Include \(x=0\)
Suppose:
| \(x\) | \(y\) |
|---|---|
| 2 | 9 |
| 5 | 18 |
First find the slope:
\[ m= \frac{18-9}{5-2} = 3. \]Now use
\[ y=3x+b. \]Substitute \((2,9)\):
\[ 9=3(2)+b. \]So
\[ b=3. \]Therefore,
\[ \boxed{y=3x+3}. \]Finding \(b\) from a Point and the Slope
If a line has slope \(m\) and passes through
\[ (x_1,y_1), \]substitute the point into
\[ y=mx+b. \]This gives
\[ y_1=mx_1+b. \]So
\[ \boxed{ b=y_1-mx_1 }. \]This is often the fastest way to recover the \(y\)-intercept value when slope and one point are known.
Example: Find \(b\)
A line has slope
\[ m=-2 \]and passes through
\[ (3,7). \]Then
\[ b=y-mx. \]Substitute:
\[ b=7-(-2)(3). \]So
\[ b=13. \]Therefore,
\[ \boxed{y=-2x+13}. \]Constructing an Equation from Two Points
Suppose a line passes through
\[ (-1,2) \]and
\[ (3,10). \]First find the slope:
\[ m= \frac{10-2}{3-(-1)} = \frac84 = 2. \]Now use one point to find \(b\):
\[ 2=2(-1)+b. \]So
\[ b=4. \]Therefore,
\[ \boxed{y=2x+4}. \]Why Two Points Are Enough
Two distinct points with different \(x\)-coordinates determine:
- the slope \(m\);
- then the intercept \(b\).
Once both parameters are known, the equation
\[ y=mx+b \]is completely determined.
Vertical Lines Are the Exception
If two points have the same \(x\)-coordinate, such as
\[ (4,-1) \]and
\[ (4,6), \]they determine the vertical line
\[ \boxed{x=4}. \]Its slope is undefined, so it cannot be written in slope-intercept form.
Every Nonvertical Line Has Slope-Intercept Form
A nonvertical line has a finite slope \(m\).
It also crosses the \(y\)-axis at some point
\[ (0,b). \]Therefore, every nonvertical line can be represented as
\[ \boxed{y=mx+b}. \]Slope-intercept form is therefore a complete representation of nonvertical straight lines in the plane.
Constant Functions in Slope-Intercept Form
A constant function
\[ y=c \]can be written
\[ y=0x+c. \]So in slope-intercept form:
\[ \boxed{m=0,\qquad b=c}. \]The graph is horizontal.
The Zero Function
The zero function is
\[ y=0. \]In slope-intercept form,
\[ y=0x+0. \]Thus,
\[ \boxed{m=0,\qquad b=0}. \]Its graph is the \(x\)-axis.
Direct Proportion as a Special Case
If
\[ b=0, \]then slope-intercept form becomes
\[ \boxed{y=mx}. \]The graph passes through the origin.
This is a direct proportional relationship, and the slope \(m\) is also the constant of proportionality.
Affine but Not Directly Proportional
Consider
\[ y=4x+9. \]The slope is constant, but
\[ y(0)=9\ne0. \]So the relationship is affine but not directly proportional.
The nonzero intercept shifts the line away from the origin.
Finding the \(x\)-Intercept from Slope-Intercept Form
For
\[ y=mx+b, \]set
\[ y=0. \]Then
\[ 0=mx+b. \]If
\[ m\ne0, \]solve:
\[ \boxed{x=-\frac{b}{m}}. \]So the \(x\)-intercept is
\[ \boxed{\left(-\frac{b}{m},0\right)}. \]Example: Find the \(x\)-Intercept
Consider
\[ y=-3x+12. \]Set \(y=0\):
\[ 0=-3x+12. \]Then
\[ 3x=12, \]so
\[ x=4. \]The \(x\)-intercept is
\[ \boxed{(4,0)}. \]Converting an Equation into Slope-Intercept Form
An equation may represent a line without already being written as
\[ y=mx+b. \]To expose the slope and \(y\)-intercept, solve the equation for \(y\).
Example: Convert from Standard Form
Rewrite
\[ 3x+2y=10 \]in slope-intercept form.
Subtract \(3x\):
\[ 2y=-3x+10. \]Divide by \(2\):
\[ \boxed{ y=-\frac32x+5 }. \]Now we can read:
\[ \boxed{m=-\frac32} \]and
\[ \boxed{b=5}. \]Example with Negative \(y\)-Coefficient
Rewrite
\[ 4x-5y=20. \]Subtract \(4x\):
\[ -5y=-4x+20. \]Divide by \(-5\):
\[ \boxed{ y=\frac45x-4 }. \]Therefore,
\[ m=\frac45, \qquad b=-4. \]Watch the Signs When Dividing
In
\[ -5y=-4x+20, \]dividing by \(-5\) changes both terms on the right:
\[ \frac{-4x}{-5} = \frac45x, \]and
\[ \frac{20}{-5} = -4. \]Sign errors during this step can change both slope and intercept.
When Standard Form Cannot Become Slope-Intercept Form
Consider
\[ 3x=12. \]This simplifies to
\[ x=4. \]There is no \(y\)-term to solve for.
The equation represents a vertical line, so slope-intercept form does not apply.
Checking a Slope-Intercept Equation
Suppose we claim that the line through
\[ (2,7) \]with slope
\[ 3 \]has equation
\[ y=3x+1. \]Check the point:
\[ 3(2)+1=7. \]Check the slope:
\[ m=3. \]Both conditions are satisfied, so the equation is correct.
Checking with a Second Point
For
\[ y=3x+1, \]increase \(x\) from \(2\) to \(5\):
\[ \Delta x=3. \]Since \(m=3\),
\[ \Delta y=3(3)=9. \]Starting from \(y=7\),
\[ 7+9=16. \]Check:
\[ 3(5)+1=16. \]The slope behavior and formula agree.
Slope-Intercept Form as a Transformation
Start with the parent function
\[ y=x. \]Multiplying by \(m\) gives
\[ y=mx, \]which changes the slope.
Adding \(b\) gives
\[ y=mx+b, \]which translates the graph vertically.
So slope-intercept form can be viewed structurally as:
\[ \boxed{\text{scale by }m\text{, then translate by }b}. \]Changing \(b\) While Holding \(m\) Fixed
Compare
\[ y=2x-4, \] \[ y=2x, \]and
\[ y=2x+6. \]Each has slope
\[ 2. \]Only the vertical placement changes.
The lines have the same rate of change but different \(y\)-intercepts.
Changing \(m\) While Holding \(b\) Fixed
Compare
\[ y=x+3, \] \[ y=2x+3, \]and
\[ y=-x+3. \]Each passes through
\[ (0,3). \]But their rates and directions differ because their slopes differ.
Families of Lines
The expression
\[ y=mx+b \]describes an entire two-parameter family of nonvertical lines.
Changing \(m\) changes the slope.
Changing \(b\) changes the \(y\)-intercept.
Each ordered pair of parameter values
\[ (m,b) \]selects one member of the family.
Same Slope, Different Intercepts
If two equations have the same \(m\) but different \(b\)-values, such as
\[ y=4x+1 \]and
\[ y=4x-7, \]they have the same rate of change.
Their graphs never change their vertical separation.
This observation will become important when studying parallel lines.
Same Intercept, Different Slopes
If two equations have the same \(b\) but different slopes, such as
\[ y=2x+5 \]and
\[ y=-3x+5, \]both contain
\[ (0,5). \]They therefore intersect at the shared \(y\)-intercept unless the equations are identical.
Slope-Intercept Form from a Change Rule
Suppose a quantity starts at \(12\) and increases by \(5\) for each unit of input.
The starting value gives
\[ b=12. \]The rate gives
\[ m=5. \]Therefore,
\[ \boxed{y=5x+12}. \]Example: Taxi Fare
A taxi charges a \$6 initial fee plus \$2.75 per mile.
Let \(d\) be distance in miles and \(C\) be cost in dollars.
The rate is
\[ m=2.75\frac{\text{dollars}}{\text{mile}}, \]and the starting value is
\[ b=6\text{ dollars}. \]So
\[ \boxed{C(d)=2.75d+6}. \]Interpret the Parameters in Context
For
\[ C(d)=2.75d+6, \]the slope
\[ 2.75 \]does not merely mean “the coefficient of \(d\).”
It means
\[ 2.75\text{ dollars per mile}. \]The intercept
\[ 6 \]means the cost at zero miles is \$6.
Good modeling requires both algebraic identification and contextual interpretation.
Example: Draining Tank
A tank begins with \(800\) liters and loses \(25\) liters per minute.
Let \(t\) be time in minutes.
The slope is negative:
\[ m=-25. \]The initial value is
\[ b=800. \]Therefore,
\[ \boxed{V(t)=-25t+800}. \]Why the Slope Is Negative in a Decreasing Model
In the tank model, each additional minute changes the volume by
\[ -25\text{ liters}. \]So
\[ \frac{\Delta V}{\Delta t} = -25\frac{\text{L}}{\text{min}}. \]The sign of \(m\) should agree with the physical direction of change.
Example: Temperature Conversion
The Fahrenheit temperature corresponding to Celsius temperature \(C\) is
\[ \boxed{ F=\frac95C+32 }. \]This is slope-intercept form.
The slope
\[ \frac95 \]rescales temperature differences.
The intercept
\[ 32 \]accounts for the different zero points of the two temperature scales.
Slope-Intercept Form and Units
In an applied equation
\[ y=mx+b, \]the units must be compatible.
If \(x\) is measured in hours and \(y\) in dollars, then \(m\) must have units
\[ \frac{\text{dollars}}{\text{hour}}. \]Then
\[ mx \]has units of dollars, matching \(b\) and \(y\).
Dimensional Check
Suppose someone proposes
\[ C(t)=12t+5, \]where \(C\) is measured in dollars and \(t\) in hours.
Then \(12\) should be interpreted as
\[ 12\frac{\text{dollars}}{\text{hour}}, \]so
\[ 12\frac{\text{dollars}}{\text{hour}} \cdot t\text{ hours} \]produces dollars.
The constant \(5\) must also be in dollars.
Restricted Domains
An algebraic line may extend indefinitely, but a real-world model often should not.
For the tank model
\[ V(t)=-25t+800, \]negative time is not meaningful.
Also, the model should not continue after the tank reaches zero volume.
Solve:
\[ 0=-25t+800. \]Then
\[ t=32. \]A natural model domain is therefore
\[ \boxed{0\le t\le32}. \]A Formula Can Be Correct but a Model Can Be Misused
If we substitute
\[ t=40 \]into
\[ V(t)=-25t+800, \]we obtain
\[ V(40)=-200. \]The arithmetic is correct, but a negative physical volume is not meaningful in this situation.
The domain restriction is part of the model.
Finding a Missing Parameter
Suppose
\[ y=mx+4 \]passes through
\[ (3,19). \]Substitute:
\[ 19=3m+4. \]Then
\[ 15=3m, \]so
\[ \boxed{m=5}. \]The equation is
\[ \boxed{y=5x+4}. \]Finding a Missing Intercept
Suppose
\[ y=-3x+b \]passes through
\[ (4,-7). \]Substitute:
\[ -7=-3(4)+b. \]Then
\[ -7=-12+b, \]so
\[ \boxed{b=5}. \]Therefore,
\[ \boxed{y=-3x+5}. \]Comparing Two Affine Models
Consider
\[ A(x)=4x+20 \]and
\[ B(x)=6x+8. \]Model \(B\) has the larger slope:
\[ 6>4. \]So \(B\) grows faster.
Model \(A\) has the larger initial value:
\[ 20>8. \]So \(A\) starts higher.
Slope-intercept form makes both comparisons immediate.
Where Two Models Are Equal
Using
\[ A(x)=4x+20 \]and
\[ B(x)=6x+8, \]set them equal:
\[ 4x+20=6x+8. \]Then
\[ 12=2x, \]so
\[ x=6. \]The common output is
\[ A(6)=44. \]Thus the graphs intersect at
\[ \boxed{(6,44)}. \]This previews how systems of equations compare competing affine relationships.
Reading a Formula Efficiently
When an equation is already in slope-intercept form, do not perform unnecessary algebra.
For
\[ y=-\frac73x+11, \]we can immediately state:
\[ m=-\frac73, \]and
\[ b=11. \]The \(y\)-intercept is
\[ (0,11). \]Only additional questions, such as finding the \(x\)-intercept, require more calculation.
Coefficient of \(x\) When It Is Not Written
Consider
\[ y=x-8. \]The coefficient of \(x\) is implicitly
\[ 1. \]So
\[ \boxed{m=1,\qquad b=-8}. \]Negative \(x\) Means Slope \(-1\)
Consider
\[ y=-x+6. \]The coefficient of \(x\) is
\[ -1. \]Therefore,
\[ \boxed{m=-1,\qquad b=6}. \]No Constant Term Means \(b=0\)
Consider
\[ y=7x. \]This can be written
\[ y=7x+0. \]Therefore,
\[ \boxed{m=7,\qquad b=0}. \]The line passes through the origin.
No \(x\)-Term Means \(m=0\)
Consider
\[ y=-4. \]This can be written
\[ y=0x-4. \]Therefore,
\[ \boxed{m=0,\qquad b=-4}. \]The line is horizontal.
Slope-Intercept Form and Function Notation
Instead of
\[ y=mx+b, \]we may write
\[ \boxed{f(x)=mx+b}. \]The mathematical structure is the same.
Function notation emphasizes that each allowed input \(x\) is assigned an output \(f(x)\).
Recovering the Output
Suppose
\[ f(x)=2.5x-3. \]At
\[ x=8, \]the output is
\[ f(8)=2.5(8)-3. \]So
\[ f(8)=20-3=17. \]Thus,
\[ \boxed{f(8)=17}. \]Recovering the Input
Suppose
\[ y=4x+7 \]and the output is
\[ y=31. \]Then
\[ 31=4x+7. \]Subtract \(7\):
\[ 24=4x. \]Therefore,
\[ \boxed{x=6}. \]The Difference Property
For
\[ f(x)=mx+b, \]the difference between two outputs is
\[ f(x_2)-f(x_1) = m(x_2-x_1). \]The intercept \(b\) cancels.
This explains why \(m\) controls change while \(b\) controls position.
Why Vertical Translation Does Not Change Slope
Suppose
\[ f(x)=mx \]and
\[ g(x)=mx+b. \]For any two inputs,
\[ g(x_2)-g(x_1) = (mx_2+b)-(mx_1+b). \]The \(b\)-terms cancel:
\[ = m(x_2-x_1). \]So the constant vertical shift does not change the rate of change.
Slope-Intercept Form as a Prediction Rule
In a constant-rate model,
\[ y=mx+b \]can be read as:
\[ \boxed{ \text{output} = (\text{rate})(\text{input}) + \text{starting value} }. \]This verbal structure is often the fastest route from a practical description to an equation.
Example: Hourly Pay
An employee receives a \$50 daily allowance plus \$22 per hour worked.
If \(h\) is the number of hours and \(P\) is total pay, then
\[ \boxed{P(h)=22h+50}. \]The slope is
\[ 22\frac{\text{dollars}}{\text{hour}}, \]and the intercept is
\[ 50\text{ dollars}. \]Example: Starting Distance
A vehicle is already \(15\) miles from a reference point when timing begins and then travels away at \(60\) miles per hour.
After \(t\) hours,
\[ \boxed{d(t)=60t+15}. \]The slope gives speed, while the intercept gives initial distance.
Example: Moving Toward the Reference Point
If the same vehicle begins \(15\) miles away but moves toward the reference point at \(5\) miles per hour, then
\[ \boxed{d(t)=-5t+15}. \]The negative slope indicates decreasing distance.
The \(x\)-intercept occurs when
\[ -5t+15=0, \]so
\[ t=3. \]The model reaches the reference point after \(3\) hours.
Approximate Slope-Intercept Models
Real data often do not follow one exact line.
A model may still be written approximately as
\[ y\approx mx+b. \]In that setting:
- \(m\) represents an estimated trend or average rate;
- \(b\) represents the model's estimated output at \(x=0\).
This becomes important later in regression and trend analysis.
The Intercept May Be an Extrapolation
Suppose data were collected only for
\[ 20\le x\le50. \]A fitted model may have a \(y\)-intercept at \(x=0\), but \(x=0\) lies outside the observed data range.
The value \(b\) is still algebraically part of the fitted line, but its real-world interpretation may be uncertain.
Do not automatically treat every fitted intercept as a directly observed or physically meaningful starting value.
A Reliable Reading Workflow
- Rewrite the equation as \(y=mx+b\) if necessary.
- Identify the coefficient \(m\).
- Interpret \(m\) as slope or rate of change.
- Identify the constant \(b\).
- Write the \(y\)-intercept as \((0,b)\).
- Attach units and contextual meaning when appropriate.
A Reliable Construction Workflow
- Determine the slope \(m\).
- Determine the \(y\)-intercept value \(b\).
- Substitute into \(y=mx+b\).
- Simplify signs and fractions carefully.
- Verify a known point or condition.
- Check that the equation's slope agrees with the intended rate.
A Reliable Two-Point Workflow
- Use the two points to compute \(m=\dfrac{y_2-y_1}{x_2-x_1}\).
- Confirm that \(x_1\ne x_2\).
- Substitute \(m\) and either point into \(y=mx+b\).
- Solve for \(b\).
- Write the completed slope-intercept equation.
- Check the second point.
A Reliable Conversion Workflow
- Isolate the \(y\)-term.
- Divide by the coefficient of \(y\).
- Simplify every term.
- Read the coefficient of \(x\) as \(m\).
- Read the constant as \(b\).
- If \(y\) cannot be isolated because the equation is vertical, do not force slope-intercept form.
Common Errors
Reversing \(m\) and \(b\)
In
\[ y=mx+b, \]\(m\) is the slope and \(b\) is the \(y\)-intercept value.
Writing the Intercept as a Number When a Point Is Needed
If \(b=5\), the \(y\)-intercept point is
\[ (0,5). \]Forgetting Invisible Coefficients
In
\[ y=x+4, \]the slope is \(1\).
In
\[ y=-x+4, \]the slope is \(-1\).
Assuming No Constant Term Means No Intercept
If
\[ y=3x, \]then \(b=0\), so the \(y\)-intercept is the origin.
Assuming No \(x\)-Term Means No Slope
If
\[ y=7, \]then \(m=0\). The line is horizontal.
Trying to Put a Vertical Line into \(y=mx+b\)
A vertical line has undefined slope and cannot be represented in slope-intercept form.
Losing Signs During Conversion
When dividing an equation by a negative coefficient, every term must be divided by that negative number.
Ignoring Units
In a model, \(m\) should be interpreted as output units per input unit, while \(b\) has the same units as the output.
Assuming the Intercept Is Always Meaningful in Context
A fitted or unrestricted algebraic intercept may lie outside the realistic or observed domain.
Practice
1. Identify \(m\) and \(b\)
For
\[ y=6x-11, \]identify the slope and \(y\)-intercept.
Check Answer
\[ \boxed{m=6} \]and
\[ \boxed{b=-11}. \]The \(y\)-intercept is
\[ \boxed{(0,-11)}. \]2. Invisible Coefficient
For
\[ y=-x+8, \]identify \(m\) and \(b\).
Check Answer
\[ \boxed{m=-1,\qquad b=8}. \]3. Build from Slope and Intercept
Write the equation of a line with slope
\[ m=\frac34 \]and \(y\)-intercept
\[ (0,-2). \]Check Answer
\[ \boxed{ y=\frac34x-2 }. \]4. Find \(b\) from a Point
A line has slope
\[ m=4 \]and passes through
\[ (3,17). \]Find its slope-intercept equation.
Check Answer
\[ 17=4(3)+b, \] \[ 17=12+b, \] \[ b=5. \]Therefore,
\[ \boxed{y=4x+5}. \]5. Build from Two Points
Find the slope-intercept equation through
\[ (2,3) \]and
\[ (6,11). \]Check Answer
\[ m= \frac{11-3}{6-2} = \frac84 = 2. \]Use \((2,3)\):
\[ 3=2(2)+b, \] \[ b=-1. \]So
\[ \boxed{y=2x-1}. \]6. Convert to Slope-Intercept Form
Rewrite
\[ 2x+4y=16. \]Check Answer
\[ 4y=-2x+16, \] \[ \boxed{ y=-\frac12x+4 }. \]Thus,
\[ m=-\frac12, \qquad b=4. \]7. Constant Function
Write
\[ y=9 \]in slope-intercept form and identify \(m\) and \(b\).
Check Answer
\[ y=0x+9. \]Therefore,
\[ \boxed{m=0,\qquad b=9}. \]8. Direct Proportion
Is
\[ y=-7x \]directly proportional? Identify \(m\) and \(b\).
Check Answer
Yes.
\[ \boxed{m=-7,\qquad b=0}. \]Because \(b=0\), the graph passes through the origin.
9. Interpret a Model
A subscription costs \$18 initially plus \$6 per month:
\[ C(t)=6t+18. \]Interpret \(m\) and \(b\).
Check Answer
\[ \boxed{m=6\text{ dollars per month}} \]is the monthly rate, and
\[ \boxed{b=18\text{ dollars}} \]is the initial cost at \(t=0\).
10. Find the \(x\)-Intercept
Find the \(x\)-intercept of
\[ y=5x-30. \]Check Answer
\[ 0=5x-30, \] \[ x=6. \]So the \(x\)-intercept is
\[ \boxed{(6,0)}. \]11. Missing Parameter
The line
\[ y=mx-3 \]passes through
\[ (4,9). \]Find \(m\).
Check Answer
\[ 9=4m-3, \] \[ 12=4m, \] \[ \boxed{m=3}. \]12. Vertical-Line Check
Can the line
\[ x=-5 \]be written in slope-intercept form?
Check Answer
No.
It is vertical, so its slope is undefined. There is no finite number \(m\) for which
\[ y=mx+b \]describes the line.
Key Ideas
- Slope-intercept form is \(y=mx+b\).
- \(m\) is the slope or constant rate of change.
- \(b\) is the \(y\)-intercept value, giving the point \((0,b)\).
- The form separates rate from starting value.
- Every nonvertical line can be written in slope-intercept form.
- Vertical lines cannot be written in slope-intercept form because their slopes are undefined.
- When \(b=0\), the relationship becomes \(y=mx\), a direct proportion and a structural linear function.
- When \(m=0\), the equation represents a constant horizontal line.
- A line can be graphed efficiently by plotting \((0,b)\) and then using \(m\) as rise over run.
- Given \(m\) and one point, the intercept can be found from \(b=y-mx\).
- Given two points with different \(x\)-coordinates, first find \(m\), then solve for \(b\).
- Other line equations can often be converted to slope-intercept form by solving for \(y\).
- For \(m\ne0\), the \(x\)-intercept is \(\left(-\frac{b}{m},0\right)\).
- In applications, \(m\) carries output units per input unit, while \(b\) has output units.
- Real-world domains may restrict how far an affine model should be interpreted.
Where This Leads
Slope-intercept form gives a compact and highly readable description of a nonvertical line. It combines the rate information developed through slope with the positional information supplied by the \(y\)-intercept, making it especially useful for graphing, modeling, comparison, and quick interpretation.
Next, the same straight-line structure can be written from a known point without first finding the \(y\)-intercept. This leads naturally to point-slope form and two-point form, followed by standard form, special line relationships, graphing methods, and constructing equations from increasingly varied conditions.