The parent function for linear functions, slope-intercept form and the four-step graphing procedure, reading slope and intercept as a rate and an initial value, standard form and the intercept method, and the equations of horizontal and vertical lines.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 2 — Linear Equations and Functions
Graph Equations of Lines
Objectives
Five outcomes. The second replaces the table of values you used in Lesson 2.1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 89-97 — the lesson these objectives are drawn from
Warm-up
Lesson 2.1 graphed by table; Lesson 2.2 named the constant change. This lesson combines them into a method that needs no table at all.
Discussion prompt
For y equals 3x minus 2, what is the constant change per step in x, and what is the output when x is zero? What do those two numbers let you draw without computing anything else?
Hint: Both numbers are already visible in the equation.
Answer:
\[ y = 3x - 2 \;\Longrightarrow\; \text{change per step} = 3, \quad y(0) = -2 \]
A point and a direction. Plot the point where the line crosses the vertical axis, then step three up and one across to find a second point. Two points determine a line, so the table was never necessary — it was only a slow way of finding those two numbers.
Concept
The family of linear functions has one parent: f of x equals x. Every other member is that same line with its steepness changed by m and its height changed by b. Recognising a family and its parent is a habit this book uses for every function type it meets.
parent function — The most basic function in a family of functions. For linear functions it is f of x equals x, whose slope is 1 and whose y-intercept is 0.
\[ y = mx + b \]
The y-intercept is the y coordinate where the graph crosses the vertical axis. Some books define it as the point itself; either way the number b is what you plot.
Figure (svg): Three lines on one plane: the parent function y equals x, a steeper line y equals 2x, and a shifted line y equals x plus 3
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 89-90
Section
Section 1
Concept
Changing m tilts the line about its intercept; changing b slides the whole line up or down without changing its direction. Comparing any linear graph with the parent is a matter of naming which dial moved.
y-intercept — The y-coordinate of the point where a graph crosses the vertical axis. For y equals m x plus b it is b, because setting x to zero leaves y equal to b.
That the intercept is b is not a coincidence to memorise: substituting x equal to zero into m x plus b leaves b, and x equal to zero is exactly what the vertical axis means.
Figure (svg): Three lines on one plane: the parent function y equals x, a steeper line y equals 2x, and a shifted line y equals x plus 3
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 89-89 — Parent Function for Linear Functions
Picture it
Example 1: the parent, a steeper line, and a slid line.
Figure (svg): Three lines on one plane: the parent function y equals x, a steeper line y equals 2x, and a shifted line y equals x plus 3
The steeper line still passes through the origin, so only its slope changed. The slid line still rises at forty-five degrees, so only its intercept changed. Every linear function is some combination of those two moves.
Worked example
Example 1, both parts. The comparison is the answer, not the picture.
\[ \text{Graph } y = 2x \text{ and } y = x + 3, \text{ and compare each with } y = x. \]
Write each in the form m x plus b
Why: The first is 2x plus 0; the second is 1x plus 3. Writing the invisible parts makes the comparison mechanical.
\[ y = 2 x + 0\text{ and } y = 1 x + 3 \]
Compare the first with the parent
Why: Both have a y-intercept of zero, so both pass through the origin; the slope changed from 1 to 2.
Compare the second with the parent
Why: Both have slope 1, so they are equally steep; the intercept changed from 0 to 3.
Say what each change does to the picture
Why: A larger slope tilts the line steeper; a larger intercept slides it upward.
Figure (svg): The solution to Worked example compare two lines with the parent shown as a ladder of expressions, one row per algebraic move
\[ y = 2x: \text{ slope } 2 \text{ instead of } 1 \qquad y = x + 3: \text{ intercept } 3 \text{ instead of } 0 \]
Verify: test one point on each line
Why: On y equals 2x, the point (1,2) works and (1,1) does not, so it really is steeper than the parent at x equal to 1. On y equals x plus 3, the point (1,4) works — exactly three above the parent's (1,1) — which is what sliding up by three means at every x.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 89-89
Sorting
Compare each with the parent y equals x, and say which dial moved.
Sort into buckets
Sort each equation by what changed from the parent.
Worked example
Guided Practice 1 and 2, done as one comparison.
\[ \text{Compare } y = -2x \text{ and } y = x - 2 \text{ with } y = x. \]
Write both with the invisible parts shown
Why: Negative two x plus zero, and one x minus two.
\[ y = -2 x + 0, y = 1 x - 2 \]
Compare the first
Why: Same intercept of zero, but the slope is negative two rather than one — so it is steeper AND it falls rather than rises.
Compare the second
Why: Same slope of one, so equally steep and still rising, but slid DOWN two rather than up.
Note the two independent changes
Why: Sign of the slope changes direction; size of the slope changes steepness; the intercept changes height. Three effects, two dials.
Figure (svg): The solution to Worked example a negative slope and a downward slide shown as a ladder of expressions, one row per algebraic move
\[ y = -2x: \text{ falls, twice as steep} \qquad y = x - 2: \text{ same tilt, slid down } 2 \]
Verify: check the point at x equal to 1 on each
Why: On y equals negative two x the point (1,-2) works, which is below the axis where the parent is at (1,1) — a fall, as a negative slope requires. On y equals x minus 2 the point (1,-1) works, exactly two below the parent's (1,1).
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 90-90
Trap
\[ \text{Compare } y = x + 3 \text{ with } y = x. \]
Say the line is steeper, because the numbers are bigger
Why: Any change in the equation is read as a change in steepness.
But both lines rise one unit for every unit across. They are exactly as steep as each other.
\[ \text{Compare } y = x + 3 \text{ with } y = x. \]
Compare the two numbers in m x plus b separately
Why: The slope 1 is unchanged, so the tilt is unchanged; the intercept moved from 0 to 3, so the line slid up.
\[ y = 1x + 3 \quad \text{versus} \quad y = 1x + 0 \]
The two lines are parallel — equal slopes, as Lesson 2.2 defined it. Parallel lines look different on a page and are equally steep.
Matching
Each is described relative to the parent line.
Match the pairs
Why: The third is worth pausing on: a slope of negative one has the same magnitude as the parent's slope of one, so the line is exactly as steep, just mirrored. Steepness is the size of the slope; direction is its sign. Keeping those two apart is what makes descriptions like this precise.
Tweak it
Change m and b and watch which feature of the line responds.
Parameter explorer
Which change moves the line without tilting it, and which tilts it without moving where it crosses the vertical axis?
\[ y = {m}x + {b} \]
Explain it to yourself
The rule is stated so often that its reason gets skipped.
\[ y = mx + b \]
Discussion prompt
Explain why the constant term of a slope-intercept equation is always where the line crosses the vertical axis. What is special about the points on that axis, and what happens to the m x term there?
Hint: What is the x coordinate of every point on the vertical axis?
Answer:
Every point on the vertical axis has x equal to zero. Substituting zero makes the m x term vanish, whatever m is, leaving y equal to b.
\[ y = m(0) + b = b \]
So the crossing point is (0, b), and it is unaffected by the slope — which is exactly why changing m pivots the line about that point rather than moving it.
Section
Section 2
Concept
When an equation is written as y equals m x plus b, the number b gives you a point to start from and the number m tells you how to step to the next one. Four steps and the line is drawn.
slope-intercept form — The form y equals m x plus b, in which m is the slope of the line and b is its y-intercept.
\[ y = -\tfrac{2}{3}x - 1 \]
A negative slope can be stepped either way: down two and right three, or up two and left three. Both land on the same line, which is a useful freedom when the first choice runs off the page.
Figure (svg): The four steps of graphing from slope-intercept form applied to y equals negative two thirds x minus one
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 90-90 — Using Slope-Intercept Form to Graph an Equation
Picture it
Example 2: y equals negative two thirds x minus one.
Figure (svg): The four steps of graphing from slope-intercept form applied to y equals negative two thirds x minus one
Solve for y, plot the intercept, step off the slope, join. The only step that ever takes effort is the first, and only when the equation arrives in another form.
Worked example
Example 2, all four steps written out.
\[ \text{Graph } y = -\tfrac{2}{3}x - 1. \]
Check the equation is solved for y
Why: It is, so step one costs nothing here.
\[ \text{already } y =\text{ mx } +b \]
Identify b and plot the point (0, b)
Why: The constant term is negative one, so the line crosses the vertical axis at negative one.
\[ \text{plot } (0, -1) \]
Identify m and step to a second point
Why: Negative two thirds means down two for every three across, so from (0,-1) go down 2 and right 3.
\[ \text{second point } (3, -3) \]
Draw the line through the two points
Why: Two points determine the line; a third is optional insurance.
Figure (svg): The solution to Worked example graph from slope-intercept form shown as a ladder of expressions, one row per algebraic move
\[ y = -\tfrac{2}{3}x - 1 \]
Verify: step the other way and land on the line
Why: From (0,-1) going up 2 and left 3 gives (-3, 1). Substituting: negative two thirds of negative three is positive two, minus one is one — correct. Both directions produce genuine points, which confirms that a negative slope may be stepped either way.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 90-90
Fill the middle
Exercise 25 in disguise: get it into slope-intercept form.
Fill in the blanks
3x - 2y = 12 \;\longrightarrow\; -2y = 12 - 3x \;\longrightarrow\; y = \frac{3}{2}x - 6
Why: Dividing 12 by negative two gives negative six, and dividing negative 3x by negative two gives positive three halves x. Both terms change sign because the divisor is negative — the trap from Lesson 1.4, met again. The line therefore rises with slope three halves and crosses the vertical axis at negative six.
Worked example
Exercise 23. The equation arrives in the wrong form, so step one earns its place.
\[ \text{Write } 4x - 3y = 18 \text{ in slope-intercept form, then describe its graph.} \]
Subtract 4x from each side
Why: Isolating the y term is the first move, exactly as in Lesson 1.4.
\[ -3 y = 18 - 4 x \]
Divide every term by -3
Why: Dividing by a negative changes the sign of every term on the other side.
\[ y = -6 + (\frac{4}{3}) x \]
Write it in the standard order
Why: The x term first, then the constant, so that m and b are where the eye expects them.
\[ y = (\frac{4}{3}) x - 6 \]
Read off the slope and intercept
Why: Slope four thirds, intercept negative six: start at (0,-6) and step up 4, right 3.
\[ m = \frac{4}{3}, b = -6 \]
Figure (svg): The solution to Worked example rearrange, then graph shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{4}{3}x - 6 \]
Verify: substitute a point into the ORIGINAL equation
Why: Stepping from (0,-6) up 4 and right 3 gives (3,-2). In the original: 4 times 3 minus 3 times negative 2 is 12 plus 6, which is 18 — the right side exactly. Checking against the original rather than the rearranged form is what validates the division by negative three.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 93-93
Trap
\[ \text{Graph } y = 2x + 3. \]
Start at the origin and step up 2, right 1
Why: The slope is applied from where the parent line starts rather than from this line's own intercept.
That produces the point (1,2), which is on y equals 2x, not on y equals 2x plus 3.
\[ \text{Graph } y = 2x + 3. \]
Plot the intercept FIRST, then step from there
Why: The slope describes the direction of the line; the intercept says where the line is. Both are needed and the intercept comes first.
\[ (0, 3) \;\longrightarrow\; (1, 5) \]
Substituting x equal to 1 gives 5, confirming (1,5). The step is right; it was the starting point that was wrong.
Elimination
Exercise 23, as it appears in the book.
Eliminate the wrong options
What is the slope-intercept form of 4x - 3y = 18?
Survives elimination: A
Why: Subtracting 4x gives -3y = 18 - 4x, and dividing every term by -3 gives y = -6 + (4/3)x. The quickest check is the intercept: setting x to zero in the original gives -3y = 18, so y = -6, which only one option matches.
Discrimination
Do not graph anything. Say which equations are ready to plot immediately.
Sort into buckets
Sort each equation by whether it is already in slope-intercept form.
Reverse engineer
A line crosses the vertical axis at 4 and falls two units for every five across.
Fill in the blanks
y = -\frac{2}{5}x + 4
Why: The intercept is given directly as 4, so only the slope has to be worked out: a fall of two over a run of five is negative two fifths. Reading an equation off a graph is the reverse of the four-step procedure, and it is what Lesson 2.4 turns into a general method for any two points.
Section
Section 3
Concept
When a line models a real situation, its y-intercept is the value at the start, when the input is zero, and its slope is the average rate of change with units. Reading a model means naming both.
\[ y = 5x + 42 \]
This is the same rate-of-change idea from Lesson 2.2, now visible in the equation rather than computed from two measurements.
Figure (svg): A line modelling a walrus calf's body length against age, with the y-intercept marked as the newborn length and the slope as growth per month
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 91-91 — the walrus calf model
Picture it
Example 3: a walrus calf's body length in inches, against its age in months.
Figure (svg): A line modelling a walrus calf's body length against age, with the y-intercept marked as the newborn length and the slope as growth per month
Forty-two inches at zero months is a newborn; five inches per month is the growth rate. Both numbers were in the equation before the graph was drawn.
Worked example
Example 3, all three parts.
\[ \text{A walrus calf's length is } y = 5x + 42 \text{ inches at age } x \text{ months. Interpret and estimate at } x = 10. \]
Identify the y-intercept and say what it means
Why: Forty-two is the length when the age is zero, which is a newborn calf.
\[ 42\text{ in at birth} \]
Identify the slope and attach its units
Why: Five is inches gained per month, because y is in inches and x in months.
\[ 5\text{ inches per month} \]
Graph using the intercept and the slope
Why: Start at (0,42) and step up 5 for each 1 across.
\[ \text{line from } (0, 42) \]
Read or compute the value at ten months
Why: Five times ten is fifty, plus forty-two.
\[ y = 92 \]
Figure (svg): The solution to Worked example interpret and estimate shown as a ladder of expressions, one row per algebraic move
\[ y(10) = 5(10) + 42 = 92 \text{ inches} \]
Verify: check the answer against the graph and the units
Why: Reading the graph at ten months gives a value just above ninety, matching the computed 92. The units also work: inches per month times months gives inches, added to the 42 inches of the newborn length. A calf growing from 42 to 92 inches in under a year is plausible for a walrus.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 91-91
Matching
Four models, each with an intercept and a slope worth naming.
Match the pairs
Why: The last one has an intercept of zero, which is meaningful rather than absent: at time zero the journey has covered no distance. A zero intercept says the model passes through the origin, which is what makes it a direct variation — the subject of Lesson 2.5.
Worked example
The paramotor model from Lesson 1.5, now read as a line.
\[ \text{A descent is modelled by } h = 2000 - 250t \text{ feet after } t \text{ minutes. Interpret and find } h \text{ at } t = 7. \]
Write it in slope-intercept order
Why: Negative two hundred and fifty t plus two thousand — the terms may be written either way round.
\[ h = -250 t + 2000 \]
Identify the intercept and its meaning
Why: Two thousand feet at time zero, which is the height at which the descent began.
\[ 2000 \text{ft}\text{ initially} \]
Identify the slope and its meaning
Why: Negative two hundred and fifty feet per minute: the negative sign is the descent.
\[ -250 \text{ft}\text{ per minute} \]
Evaluate at seven minutes
Why: Two hundred and fifty times seven is 1750, and 2000 minus 1750 is 250.
\[ h = 250 \text{ft} \]
Figure (svg): The solution to Worked example interpret a falling model shown as a ladder of expressions, one row per algebraic move
\[ h(7) = 2000 - 250(7) = 250 \text{ feet} \]
Verify: find where the model reaches zero and check it is sensible
Why: Setting h to zero gives t equal to 8 minutes, which is the landing. Seven minutes is one minute before that, and one minute of descent at 250 feet per minute is exactly the 250 feet the model gives. The two routes agree, and the model's domain is confirmed as zero to eight minutes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 91-91
Error analysis
A student is asked what the y-intercept represents in the walrus model and answers with a number alone.
Annotate
On: \( y = 5x + 42 \;\Longrightarrow\; \text{the y-intercept is } 42 \)
In a modelling question the units and the meaning are part of the answer, not decoration on it.
Real world
A pool contains 500 gallons and is draining at 12 gallons per minute.
Discussion prompt
Write the model, name what the slope and the intercept represent, and say when the pool is empty. Then say what the model claims at 50 minutes and whether you should believe it.
Hint: Draining means the amount decreases, so the slope is negative.
Answer:
\[ V = 500 - 12t \]
The intercept 500 is the initial volume in gallons; the slope negative 12 is the drainage rate in gallons per minute. Setting V to zero gives t equal to about 41.7 minutes, which is when the pool empties.
At 50 minutes the model gives negative 100 gallons, which is not a volume. The model's domain is zero to about 41.7 minutes; beyond that it is arithmetic without meaning, the same boundary you found for the paramotor in Lesson 1.5.
Estimation
The walrus model, y equals 5x plus 42.
Predict first
Roughly how long until the calf reaches 100 inches?
Correct: About 12 months.
\[ 5x + 42 = 100 \;\Longrightarrow\; 5x = 58 \;\Longrightarrow\; x = 11.6 \]
Why: The calf needs to gain 58 inches from its newborn length of 42, and at 5 inches a month that takes a bit under 12 months. Solving exactly gives 11.6 months. Estimating first is a check on the arithmetic, and it also flags whether the answer is inside the range where the model can be trusted — a walrus does not grow linearly for ever.
Socratic
One question, and nothing else on this slide.
\[ y = 5x + 42 \]
Discussion prompt
The model says a newborn calf is 42 inches long. What would it mean for the model if a real newborn were measured at 38 inches, and would that make the model useless? What would you check before deciding?
Hint: Distinguish between a model being exact and being useful.
Answer:
It would mean the model's intercept is off by four inches, which is a real discrepancy but not necessarily a fatal one. Models are fitted to data and rarely pass exactly through any single measurement.
What to check: whether the SLOPE still describes the growth well over the ages you care about. A model can have a slightly wrong starting value and still predict growth accurately, and a model can pass exactly through the first point and drift badly afterwards.
Lesson 2.6 makes this precise by fitting a line to scattered data, where no single point is expected to sit on the line at all.
Section
Section 4
Concept
In standard form both variables sit on the same side. Setting y to zero finds where the line crosses the horizontal axis; setting x to zero finds where it crosses the vertical one. Two points determine the line.
x-intercept — The x-coordinate of the point where a graph crosses the horizontal axis. It is found by substituting zero for y and solving.
\[ 5x + 2y = 10 \]
Every linear equation can be written in standard form, including vertical lines — which is the one thing slope-intercept form cannot do.
Figure (svg): The line five x plus two y equals ten drawn through its two intercepts, with the substitutions that found them
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 92-92 — Graph an equation in standard form
Picture it
Example 4: five x plus two y equals ten.
Figure (svg): The line five x plus two y equals ten drawn through its two intercepts, with the substitutions that found them
The method is quick precisely because the axes are where a variable is zero. That is what an axis means, and it is why the substitutions are so simple.
Worked example
Example 4, all four steps.
\[ \text{Graph } 5x + 2y = 10. \]
Confirm the equation is in standard form
Why: Both variables on the left, a constant on the right.
Set y to zero to find the x-intercept
Why: Five x plus zero is ten, so x is two. The line crosses the horizontal axis at (2,0).
\[ \text{x-intercept } 2 \]
Set x to zero to find the y-intercept
Why: Zero plus two y is ten, so y is five. The line crosses the vertical axis at (0,5).
\[ \text{y-intercept } 5 \]
Draw the line through the two intercepts
Why: Two points determine the line, and these two are as easy to find as any.
\[ \text{line through } (2, 0)\text{ and } (0, 5) \]
Figure (svg): The solution to Worked example graph by intercepts shown as a ladder of expressions, one row per algebraic move
\[ x\text{-intercept } 2, \quad y\text{-intercept } 5 \]
Verify: convert to slope-intercept form and compare
Why: Solving for y gives y equals negative five halves x plus five, whose intercept is 5 — matching. The slope of negative five halves also matches the picture: from (0,5) going down five and right two lands on (2,0), the other intercept. Two independent routes to the same line.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 92-92
Comparison
Fill the blanks. Each form makes something easy that the other does not.
Comparison matrix
| Feature | Slope-intercept form | Standard form |
|---|---|---|
| Shape | y = mx + b | Ax + By = C |
| What you read off directly | slope and y-intercept | neither - both need a substitution |
| Fastest way to graph | plot (0,b), step off the slope | find both intercepts |
| Can it write a vertical line? | no | yes |
The last row is the reason standard form survives at all. Every line can be written in it, which is not true of slope-intercept form.
Worked example
Guided Practice 12. The intercepts land on opposite sides of the origin.
\[ \text{Graph } 3x - 2y = 12. \]
Set y to zero
Why: Three x equals twelve, so x is four.
\[ \text{x-intercept } 4 \]
Set x to zero
Why: Negative two y equals twelve, so y is negative six.
\[ \text{y-intercept } -6 \]
Plot both points
Why: One on the positive horizontal axis, one on the negative vertical axis.
\[ (4, 0)\text{ and } (0, -6) \]
Draw the line and check its direction
Why: From (0,-6) to (4,0) the line rises, so its slope is positive — consistent with the two intercepts having opposite signs.
Figure (svg): The solution to Worked example intercepts with a negative shown as a ladder of expressions, one row per algebraic move
\[ x\text{-intercept } 4, \quad y\text{-intercept } -6 \]
Verify: compute the slope from the two intercepts
Why: From (0,-6) to (4,0) the rise is 6 and the run is 4, so the slope is three halves. Solving the original for y gives y equals three halves x minus six, whose slope is also three halves. The intercept method and the algebra agree.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 92-92
Error analysis
A student graphs 2x plus 5y equals 10 and plots the wrong two points.
Annotate
On: \( 2x + 5y = 10 \;\Longrightarrow\; \text{plot } (2, 0) \text{ and } (0, 5) \)
The check takes one substitution: put your claimed x-intercept back into the equation with y equal to zero and confirm the two sides agree.
Fill the middle
Exercise 30, from the exercise set.
Fill in the blanks
5x - 6y = 30, \; y = 0 \;\Longrightarrow\; 5x = 30 \;\Longrightarrow\; x = 6
Why: Setting y to zero removes the second term entirely, leaving 5x equal to 30 and so x equal to 6. Note the coefficient of x is 5 and the constant is 30, and the intercept is neither of them — it is their quotient. That division is what the swapped-intercepts error skips.
Sorting
Both methods work on every line. Say which is quicker for each equation as written.
Sort into buckets
Sort each equation by the faster graphing route.
Neither method is better in general. The faster one is whichever matches the form the equation arrived in.
Prediction
Commit before substituting.
Predict first
For x plus 5y equals negative 15, what are the two intercepts?
Correct: x-intercept -15, y-intercept -3.
\[ y = 0: \; x = -15 \qquad x = 0: \; 5y = -15, \; y = -3 \]
Why: Setting y to zero leaves x equal to negative fifteen, since the coefficient of x is one. Setting x to zero gives 5y equal to negative fifteen, so y is negative three. Both intercepts are negative, which means the line crosses both axes on the negative side and therefore passes below and to the left of the origin.
Section
Section 5
Concept
The graph of y equals c is the horizontal line through (0, c): every point on it has y equal to c and x may be anything. The graph of x equals c is the vertical line through (c, 0), and the roles are swapped.
\[ y = c \text{ is horizontal} \qquad x = c \text{ is vertical} \]
A vertical line cannot be written in slope-intercept form, because its slope is undefined. It can always be written in standard form, with the coefficient of y equal to zero.
Figure (svg): One plane showing the horizontal line y equals 2 and the vertical line x equals negative 3, each with a point marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 92-92 — Horizontal and Vertical Lines
Picture it
Example 5: y equals 2 and x equals negative 3.
Figure (svg): One plane showing the horizontal line y equals 2 and the vertical line x equals negative 3, each with a point marked
The equation names one coordinate and says nothing about the other, so the graph collects every point with that coordinate — which is a whole line.
Worked example
Example 5, both parts.
\[ \text{Graph } y = 2 \text{ and } x = -3. \]
Read what the first equation constrains
Why: It fixes y at 2 and says nothing about x, so x is free to be anything.
Draw every point with y equal to 2
Why: Those points form a horizontal line through (0,2).
\[ \text{horizontal through } (0, 2) \]
Read what the second equation constrains
Why: It fixes x at negative 3 and leaves y free.
Draw every point with x equal to negative 3
Why: Those points form a vertical line through (-3,0).
\[ \text{vertical through } (-3, 0) \]
Figure (svg): The solution to Worked example graph two one-variable equations shown as a ladder of expressions, one row per algebraic move
\[ y = 2: \text{ horizontal} \qquad x = -3: \text{ vertical} \]
Verify: test two points on each
Why: On the first, (5,2) and (-100,2) both have y equal to 2 and both satisfy the equation, while (5,3) does not. On the second, (-3,7) and (-3,-40) both satisfy x equal to negative 3. Every point on each line shares the named coordinate, which is what the equation asserts.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 92-92
Two truths and a lie
All three are about horizontal and vertical lines.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: C
Why: The survivor is the false one. A vertical line has an undefined slope, so there is no m to put into y equals m x plus b — and the equation cannot even be solved for y, because y does not appear in it. This is the one thing slope-intercept form cannot express, and it is why the standard form is kept around.
Worked example
Guided Practice 13 and 14, extended to show what standard form can do.
\[ \text{Write } x = 1 \text{ and } y = -4 \text{ in the form } Ax + By = C. \]
Write the first with an explicit y term
Why: Adding zero times y changes nothing and makes the form visible.
\[ 1 x + 0 y = 1 \]
Note that B is zero here
Why: A zero coefficient on y is exactly what makes the line vertical: y never constrains anything.
\[ B = 0:\text{ vertical} \]
Write the second with an explicit x term
Why: Zero times x plus one times y equals negative four.
\[ 0 x + 1 y = -4 \]
Note that A is zero here
Why: A zero coefficient on x makes the line horizontal.
\[ A = 0:\text{ horizontal} \]
Figure (svg): The solution to Worked example write both in standard form shown as a ladder of expressions, one row per algebraic move
\[ 1x + 0y = 1 \qquad 0x + 1y = -4 \]
Verify: check which of the two can also be written as y equals something
Why: The horizontal one becomes y equals negative four, which is slope-intercept form with slope zero. The vertical one cannot: solving 1x plus 0y equals 1 for y is impossible, because y does not appear. That impossibility is the algebraic face of an undefined slope.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 92-92
Error analysis
A student is asked to graph x equals 1 and draws a horizontal line at height 1.
Annotate
On: \( x = 1 \;\Longrightarrow\; \text{a horizontal line through } (0, 1) \)
Test one point that is not the obvious one. Any point on the line you drew, other than the one you started from, will expose the swap immediately.
Matching
Four one-variable equations.
Match the pairs
Why: The named variable is fixed and the line runs in the direction of the free one. An equation naming y gives a horizontal line, because y cannot change along it; an equation naming x gives a vertical line. The sign of the constant places it on one side of the origin or the other.
Edge cases
Standard form has coefficients A and B; take each to zero.
Discussion prompt
What line does Ax plus By equals C describe when B is zero? When A is zero? And what happens if both A and B are zero — is that still a line?
Hint: Try A equals 0 and B equals 0 with C equal to 5, then with C equal to 0.
Answer:
With B equal to zero the equation is Ax equals C, so x is fixed and the line is vertical. With A equal to zero it is By equals C, so y is fixed and the line is horizontal.
With both zero the equation reads 0 equals C. If C is not zero that is false for every point, so the graph is empty — no line at all. If C is zero it is true for every point, so the graph is the whole plane.
Neither degenerate case is a line, which is why the definition of standard form requires A and B not both zero.
Explain it
A classmate keeps drawing x equals 3 as a horizontal line.
Discussion prompt
In three sentences, give them a rule that fixes this permanently, and one test they can run on their own drawing to catch the mistake before they hand it in.
Hint: The rule is about which variable the equation mentions.
Answer:
The variable the equation names is the one that is stuck; the line runs along the direction of the variable that is missing. So x equals 3 pins every point's x coordinate at 3 and lets y roam, which draws a vertical line.
The test: pick any point on the line you drew other than the obvious one, and check it satisfies the equation. On a wrongly drawn horizontal line, a point like (5,3) fails x equals 3 immediately.
Comparison
Fill the blanks. The form decides the method, not the difficulty.
Comparison matrix
| Form | What you can read off | How to graph it |
|---|---|---|
| y = mx + b | slope and y-intercept | plot (0,b), step off m |
| Ax + By = C | nothing directly | find both intercepts |
| y = c | slope 0, intercept c | horizontal line through (0,c) |
| x = c | slope undefined, no y-intercept | vertical line through (c,0) |
| a table of values | the constant difference | plot and join |
The last row is Lesson 2.1's method, and it still works. It is simply slower than reading the two numbers off an equation that already contains them.
Pattern
One routine graphs any linear equation.
Step five catches the two errors this lesson is built around: stepping the slope from the origin instead of the intercept, and reading a standard-form coefficient as an intercept.
OpenStax Algebra and Trigonometry 2e, §4.1 Linear Functions §4.1
Check
Converting to slope-intercept form. Watch the division by a negative.
Check your understanding
What is the slope-intercept form of 4x - 3y = 18?
Answer: A
Why: Subtracting 4x gives -3y = 18 - 4x, and dividing every term by -3 gives y = -6 + (4/3)x. Setting x to zero in the original confirms the intercept: -3y = 18, so y = -6.
Check
The intercept method. Do the division.
Check your understanding
What is the x-intercept of the graph of 5x - 6y = 30?
Answer: A
Why: Setting y to zero gives 5x = 30, so x = 6. Checking: 5 times 6 minus 6 times 0 is 30, which matches the right side.
Check
Interpreting a model. The units are part of the answer.
Check your understanding
In the walrus model y = 5x + 42, where y is length in inches and x is age in months, what does the 5 represent?
Answer: A
Why: The 5 is the coefficient of x, which is the slope, and the slope is the rate of change: inches gained per month. The 42 is the y-intercept, which is the newborn length.
Real world
A taxi fare is 3 dollars on pickup plus 2 dollars a mile. A rival charges a flat 15 dollars for any trip up to 10 miles.
Discussion prompt
Write each as an equation, describe each graph in words including its slope and intercept, and find the distance at which the two cost the same. Which is cheaper on a short trip, and how could you tell from the graphs without solving anything?
Hint: One of the two graphs is a horizontal line.
Answer:
\[ y = 2x + 3 \qquad y = 15 \; (0 \leq x \leq 10) \]
The first is a rising line starting at 3 dollars; the second is a horizontal line at 15 dollars, a constant function with slope zero. They meet where 2x plus 3 equals 15, which gives x equal to 6 miles.
Without solving: at zero miles the first line starts far lower, and a rising line that starts below a horizontal one must stay below it until they cross. So the metered taxi is cheaper for any trip under six miles and dearer beyond it.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does every line have both an x-intercept and a y-intercept?
Correct: No — a horizontal line other than the x-axis never crosses the horizontal axis, so it has no x-intercept.
\[ y = 2: \; \text{no } x\text{-intercept} \qquad x = -3: \; \text{no } y\text{-intercept} \]
Why: The line y equals 2 sits two units above the horizontal axis and stays there for ever, so it never meets it. Similarly a vertical line other than the y-axis has no y-intercept. Every line with a nonzero, defined slope does have both, which is why the intercept method works so often — but the two special cases are exactly the ones it fails on, and they are the ones this lesson introduced.
Explain it
They graph everything with a table of five values and it takes them ten minutes a line.
Discussion prompt
In four sentences or fewer, show them how to graph y equals m x plus b in under thirty seconds, and explain why it works. Give them the one thing to check before they start.
Hint: The check is about the form the equation is in.
Answer:
Check first that y is alone on one side; if it is not, solve for it. Then plot the point where the line crosses the vertical axis, which is the constant term, and from there step up or down by the top of the slope and across by the bottom.
It works because the constant term is the value at x equal to zero, and the slope says exactly how the line moves from any point to the next. Two points determine a line, so nothing more is needed.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For solving for y, write the division out term by term so no sign is missed. For negative slopes, write the slope as a fraction with the minus on top, then go down and right. For intercepts, remember each one needs a division, not just a glance at a coefficient. For the two special lines, the named variable is the one that is fixed. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Draw one large coordinate plane in the middle of a page. On it, graph the parent line y equals x in pencil, then in a second colour graph one line that is steeper and one that is slid up, labelling each with its equation and saying in three words what changed. Around the edges of the page, write the four steps of graphing from slope-intercept form and beside them work one example of your own from equation to picture. In one corner, take a standard-form equation of your own and find both intercepts by substitution, showing the two divisions. In another corner, draw a horizontal line and a vertical line, write their equations, and write beside each which variable is fixed and which is free. Finally, take any real situation with a starting amount and a rate, write its equation, and label the intercept and the slope on the graph with what they mean in words and units.
If the labels in the last part say only numbers, go back and add the units and the meaning. A model whose numbers are unlabelled cannot be checked by anyone, including you.
Recap
Five things, and the second one retires the table of values for good.
| If you see | Do |
|---|---|
| y alone on one side | Plot (0,b), step off m |
| Both variables on one side | Set each to zero in turn |
| Only y in the equation | Draw a horizontal line |
| Only x in the equation | Draw a vertical line |
| A real model | Name the intercept and slope in words and units |
Lesson 2.4 turns the process round: given points or a slope, write the equation rather than read it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.3 Graph Equations of Lines §2.3, pp. 89-97 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.