Last updated October 13, 2026.

Secondary Math I · Module 2 · Linear & Exponential Functions

2.5 — Making My Point

Supplement for Mathematics Vision Project (MVP) curriculum · Secondary Math I, Module 2, Section 2.5
Independently created to support students using MVP Math. Not official publisher material and not affiliated with or endorsed by the Mathematics Vision Project.

Today’s goal. Write an equation for a line using any point and its slope; show that different forms can represent the same line; and connect graphs, tables, and equations. Review linear and exponential models and prove algebraic expressions equivalent.

1. In class: A Solidify Understanding Task

One quilt pattern, two equivalent equations

IN YOUR MVP PACKETSection 2.5 · Main task · pp. 27–28 · Problem 1

Explain why Zac and Sione can write different-looking equations for the same quilt-block pattern.

Suppose a quilt design has 7 blocks in figure 1, then 13 blocks in figure 2, then 19 blocks in figure 3. Each new figure adds 6 blocks, so this is an arithmetic sequence.

Figure number n	1	2	3	4
Number of blocks	7	13	19	25

Previously, you may have written a linear rule by finding the value at input 0. To reach 7 at n=1, that starting value must be 1:

But there is another way. Begin with the known value 7 at n=1. From figure 1 to figure n, there are n−1 steps; each step adds 6 blocks:

Worked example 1 — Verify Zac and Sione agree

Start with the form using figure 1 and distribute:

The expressions are equivalent: they give the same output for every allowed n. For instance, at n=4, both give 25.

We can also interpret their pieces:

1 in is the extrapolated value at n=0; 7 in is the value we actually know at n=1. They are simply different starting points for the same pattern.

Key idea: A linear rule need not be written using its y-intercept. You can start from any known point and use the constant rate of change.

Use any known point on a line

IN YOUR MVP PACKETMain task · pp. 28–29 · Problems 2–4

Explain why two students can write different-looking equations for a single line and recognize the role of a point on the line.

In a graph, the slope is the change in output divided by the change in input. When the slope is constant, you can start at any known point on the line and move along it using the same rate of change.

Worked example 2 — Equation from two points

Suppose the line contains and . Calculate slope:

Using the point , the output at any x is

Using instead:

Distribute both expressions:

Both equations describe the same line, even though they emphasize different known points.

You can also rearrange an expression such as into . Both forms say that changing x away from −6 changes y away from 5 at a slope of −½.

Point-slope form: the general rule

IN YOUR MVP PACKETMain task · pp. 29–31 · Problems 5–11

Generalize the pattern, write equations from a point and a slope, graph them, and explain why the method works.

We have now seen that it works to start with any known point, not just the point where a graph crosses the y-axis. The general rule comes from the definition of slope.

If is a point on a nonvertical line, then for any other point on the same line,

Multiplying by and rearranging gives a formula that also works when x equals :

Point-slope form

, or equivalently . Here is the slope and is any known point on the line. The minus signs are part of the formula: for a point with a negative x-coordinate, subtracting that value becomes addition.

Worked example 3 — Write an equation from a slope and a point

A line has slope 3 and passes through . Begin with point-slope form:

In the alternative form:

Check the known point: put x=4. The right side becomes −2. If you want slope-intercept form, distribute:

Worked example 4 — A negative coordinate changes the signs

A line has slope and passes through .

Distribute: . Verify x=−3 gives y=5. The plus sign in is not optional: it comes from subtracting −3.

Graph a line from its point-slope form

To graph :

Find the known point: .
Read the slope as : move right 2 and down 3 to .
Draw the line through the points. It also passes through .

Point-slope graph: slope −3/2 and known point (2, 1). The dashed triangle shows a run of 2 and a rise of −3, giving the second point (4, −2).

Common mistake: The known point is not automatically the y-intercept. To find the y-intercept, substitute x=0 (or simplify the equation to ).

Worked example 5 — Equivalent equations for the same graph

These all represent the same line:

Check by distributing in the first equation: . Different forms can emphasize a known point, a slope, or the y-intercept without changing the graph.

2. READY — Write equations from the information given

Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.

IN YOUR MVP PACKETReady · p. 32 · Problems 1–9

Use slope-intercept form when the intercept is given; point-slope form when a point and slope are given; calculate slope first when two points are given.

When the slope and y-intercept are known

The slope-intercept form is . Simply insert the given slope m and y-intercept b.

Worked example 6 — Direct slope-intercept form

Write a line with slope and y-intercept 6.

Check: when x=0, y=6. When x increases by 4, y decreases by 3.

When the slope and another point are known

Use point-slope form directly, then simplify only if you are asked to. In particular, negative coordinates require care with subtraction.

Worked example 7 — Point-slope with two negatives

Slope −4; point . Substitute:

Answer: , or . Both are in point-slope form.

When two points are given

First calculate the slope . Then use either point in point-slope form. If the x-values match, the line is vertical and cannot be written as a function y of x; such a line is not the intended case in these exercises.

Worked example 8 — Two points to point-slope form

The line passes through and .

Using the first point, . Using the second point, . Simplify either to check: both become .

Check your work — Ready 1–9 answer key

\(y=-7x+4\).
\(y=\frac38x-3\).
\(y=16x-\frac15\).
\(y=9(x-0)-7\).
\(y=\frac23(x+6)+1\).
\(y=-5(x-4)+11\).
\(y=-3(x-2)-5\).
\(y=3(x-0)-9\).
\(y=-(x+4)+8\).

3. SET — Translate starting values and changes into graphs and rules

Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.

IN YOUR MVP PACKETSet · p. 33 · Problems 10–13

Create graphs and equations for additive and multiplicative change. Choose and label a useful scale.

In these questions, the input counts the stages after an initial value. Unless a problem says otherwise, let stage 0 have the beginning value, stage 1 follow one change, and so on. Mark the x-axis “stage n” and the y-axis with the quantity being modeled.

Words in the problem	Mathematical pattern	Equation for beginning value A
Increase by 2 units every stage	Add 2 repeatedly (linear)
Decrease by 3 units every stage	Subtract 3 repeatedly (linear)
Become 4 times as large each stage	Multiply by 4 repeatedly (exponential)
Decrease by 25% each stage	Keep 75% each stage (exponential)

Worked example 9 — Additive change and graph scale

A model starts at 7 and loses 2 units at each stage. The equation is .

n	0	1	2	3	4
f(n)	7	5	3	1	−1

Plot those five pairs using a vertical scale that includes at least −1 through 7. If the model is a sequence defined at whole-number stages, plot separate points. A line through them would represent an additional real-domain version of the same algebraic rule.

Worked example 10 — Multiplicative change

A model starts at 3 and doubles at each stage. Its equation is .

n	0	1	2	3
g(n)	3	6	12	24

A scale going to at least 24 is needed for these points; 2 or 5 units per grid line would be convenient. The repeated factor, rather than a constant additive difference, makes the model exponential.

MVP clarification — Set, Problem 11

“One-quarter smaller at each stage” means each value is 25% less than the previous value. Multiply by at every step. Starting at 16, this gives .

This is different from “one-quarter as large,” which means multiply by , giving .

For Set 10–13, “beginning value” means stage 0 under our stated convention. If the class instead starts its stage count at 1, the formula must shift the index; the pattern itself is unchanged.

Check your work — Set 10–13 answer key

Here “beginning value” is at input 0. Plots are not shown.

\(f(x)=5-3x\).
If “one-quarter smaller” means subtract \(\frac14\) each stage: \(f(x)=16-\frac14x\). If it means 25% smaller: \(f(x)=16\left(\frac34\right)^x\). The wording does not distinguish these.
\(f(x)=10^x\).
\(f(x)=-8+2x\).

4. GO — Show that algebraic expressions are equivalent

Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.

IN YOUR MVP PACKETGo · p. 34 · Problems 14–20

Use the distributive property and combine like terms to prove two expressions are equivalent; construct an equivalent expression with parentheses.

Two expressions are equivalent if they give the same value for every allowed input. For instance, and are equivalent because distributing gives .

Important: In this assignment, you are not finding the value of x that solves an equation. You are simplifying one side to show that both sides are the same expression for every x.

A reliable method

1. Distribute multiplication across every term inside parentheses. 2. When a negative sign precedes parentheses, change the sign of every term inside. 3. Combine the x-terms and number terms separately. 4. Compare the simplified expression with the other side.

GO worked Problem A — Straightforward equivalence

Show that and are equivalent.

Because it simplifies to exactly , the expressions are equivalent for every x.

GO worked Problem B — Challenging nested groups

Show that equals .

Key step: The minus sign in front of the square brackets changes both signs inside the bracket.

Create your own equivalent expression

GO worked Problem C — Build an expression with two sets of parentheses

Create an expression equivalent to that uses at least two groups and one subtraction sign.

Check by removing parentheses:

This satisfies the structural requirement and is identical to the target expression for every x.

When to check with substitution

Substituting a number for x is a useful way to catch an error, but checking just one or two input values does not prove two expressions are equivalent. The algebraic simplification does.

Worked example 11 — Detect a sign mistake

Someone claims . Substitute x=0:

Left side ; right side . Different values show the expressions are not equivalent. Correct distribution gives .

Check your work — Go 14–20 answer key

\(5x-7+2(3x+1)-10x=5x-7+6x+2-10x=x-5\).
\(24-10(2x+8)+62+7x=24-20x-80+62+7x=6-13x\).
\(2x-3(-4x-5)-13=2x+12x+15-13=14x+2\).
\(6(x+3)-5(x+3)=(6-5)(x+3)=x+3\).
\(7(2x+1)-5x-3(3x+1)=14x+7-5x-9x-3=4\).
\(12+8x-3(x+4)-4x=12+8x-3x-12-4x=x\).
Sample: \(2(x-5)-(x+3)=2x-10-x-3=x-13\).

Additional Learning Resources

Resource & direct link	What it teaches / why use it	Scope / What to skip
Paul's Online Math Notes — Lines	Written examples for finding slope from two points and using point-slope and slope-intercept equations; helpful for the main task and Ready.	Some extra material. Read the slope and point-slope examples. Skip parallel/perpendicular lines and later topics for this lesson.
Khan Academy — Point-slope form review	Short explanations, examples, and practice for writing a line's equation from a point and a slope.	Close match. Especially useful for Ready Problems 4–9.
Professor Leonard — Using the Point-Slope Equation of a Line (TTP Video 45)	Optional lecture-style explanation of writing a line's equation when given a point and slope.	Close match. Useful extended explanation; specific timestamps were not verified.
Professor Leonard — Discovering Point-Slope Form (TTP Video 44)	Optional lesson deriving point-slope form and explaining why it works from the slope definition.	Close match. Directly supports the main task's conceptual goal; specific timestamps were not verified.
Khan Academy — Combining Like Terms Review	Simple explanations and practice for combining variable terms and handling negative coefficients, useful for Go Problems 14–20.	Close match (prerequisite review). Use the variable-expression examples; skip unrelated later algebra topics.
OpenStax — Chapter 4 Key Concepts	A compact reference for graphing and writing equations of lines.	Some extra material. Read the slope and point-slope summaries; skip inequalities and parallel/perpendicular exercises.

Teacher's Review of the Original MVP Packet

The main task develops a valid and valuable principle: a line can be written using its slope and any of its points, not only the y-intercept. Its multiple representations are mathematically equivalent.

Curriculum Coverage

MVP assignment	Required skills	Where taught here
Main Problem 1, pp. 27–28	Equivalent arithmetic-sequence descriptions; interpret starting point	In class, example 1
Main Problems 2–4, pp. 28–29	Derive slope from a graph; construct equivalent equations from different points	In class, example 2
Main Problems 5–11, pp. 29–31	Point-slope form, slope formula, graphing and verifying equations	In class, examples 3–5 and graph
Ready 1–3, p. 32	Slope-intercept form	Section 2, example 6
Ready 4–6, p. 32	Point-slope form given slope and point	Section 2, example 7
Ready 7–9, p. 32	Calculate slope from two points; write equation	Section 2, example 8
Set 10–13, p. 33	Starting value, linear versus exponential stage patterns, graphs and scales	Section 3, examples 9–10 and Problem 11 clarification
Go 14–19, p. 34	Distributive property; equivalent expressions with nested groups	Section 4, Go examples A–B and example 11
Go 20, p. 34	Construct and prove an equivalent expression with parentheses	Section 4, Go example C

Page numbers refer to pages printed in the MVP packet, not PDF viewer page numbers. Worked examples are original; they demonstrate skills rather than reproduce the packet answer key.

Last updated October 13, 2026.