Slope and starting point: two numbers describe any straight line.
Every linear relationship - savings over time, feed costs per day, riding lesson revenue - can be captured in one compact equation: $y = mx + b$. The slope $m$ tells you how fast things change. The y-intercept $b$ tells you where they start. Together, they let you draw a precise picture of any linear situation.
Slope-intercept form:
$$y = mx + b$$
$m$ = slope (rate of change - rise per unit of run) $b$ = y-intercept (value of $y$ when $x = 0$; the starting value)
To graph: (1) Plot the y-intercept $(0, b)$. (2) From that point, apply slope: go right $1$ unit, go up $m$ units (or down if $m < 0$). (3) Draw the line.
To write the equation: Identify $m$ (slope) and $b$ (starting value) from the context, then substitute.
Start at bName the directionMove by rise/run
Worked Example
Worked Example - Word problem to equation to graph
Mia starts with €50 in savings and saves €12 per week. Write the equation for her total savings $y$ after $x$ weeks, and describe the graph.
Starting value (y-intercept): $b = 50$ (she already has €50).
Rate of change (slope): $m = 12$ (adds €12 each week).
$$y = 12x + 50$$
Graph description: Starts at $(0, 50)$. Rises $12$ units for each $1$ unit to the right. At $x = 4$: $y = 12(4)+50 = 98$.
Warm-Up - Describe and use equations (equations given)
Problem 1
For $y=3x+7$: (a) State the slope $m$. (b) State the y-intercept $b$. (c) Name the point where the line crosses the y-axis. (d) Calculate $y$ when $x=5$. (e) Choose: the line RISES / FALLS as $x$ increases.
$m = 3$, $b = 7$
read off m and b → substitute x=5 → direction →
Problem 2
For $y=-2x+10$: (a) State the slope. (b) State the y-intercept. (c) Calculate $y$ when $x=0$. (d) Calculate $y$ when $x=3$. (e) Calculate $y$ when $x=5$. (f) Calculate the value of $x$ when $y=0$.
$m = -2$, $b = 10$
substitute each x → solve for x-intercept →
Problem 3
Mia's stable has a fixed monthly cost of €150 (rent and utilities) plus €8 per horse per day for feed. Write the equation for total monthly cost $y$ in terms of number of days $x$ the stable operates, if there are 3 horses.
$y = (\text{feed cost per day}) \times x + (\text{fixed cost})$
find m (feed/day × horses) → b = 150 → write equation →
Core Problems - write equation from context, then use it
Problem 4
The arithmetic sequence $23, 18, 13, 8, \ldots$ has $a_1=23$. The terms decrease. (a) Calculate the signed common difference $d$, then write the equation for the $n$-th term $y$ in terms of position $x$, using $y=dx+(a_1-d)$. (b) Verify the equation when $x=1$ and $x=3$. (c) State the graph's slope and explain how it relates to $d$.
general term y = dx + (a₁ - d) → verify → slope = d →
Problem 5
A line has slope $\frac{3}{2}$ and passes through $(4,11)$. (a) Calculate the y-intercept $b$. (b) Write the full equation.
11 = (3/2)(4) + b → solve for b → write equation →
Problem 6
A line passes through $(2,5)$ and $(6,13)$. (a) Calculate $\Delta y$. (b) Calculate $\Delta x$. (c) Calculate the slope $m$ with its graph units. (d) Use one point to calculate $b$. (e) Write the equation. (f) Calculate $y$ when $x=10$.
Δy → Δx → divide → substitute for b → equation → extend →
From here - write the equation yourself, then solve
Problem 7
Mia's savings account has €50 already and grows by €12 per week. A second account has €200 and grows by €5 per week. (a) Write the first account's equation. (b) Write the second account's equation. (c) Calculate when the balances are equal.
write y = 12x + 50 and y = 5x + 200 → set equal → solve for x →
Problem 8 Challenge
Create two different lines through $(4,20)$. (a) Choose the first slope and calculate its y-intercept. (b) Choose a different slope and calculate its y-intercept. (c) Answer: If two lines have equal slopes but different y-intercepts, do they cross? Explain.
substitute (4,20) → solve for b with each slope → parallel lines discussion →
Problem 9 Open
Design a business scenario with an upfront cost and positive earnings over time. (a) Describe the scenario and name what both axes measure. (b) Write a net-profit equation with a negative starting value and positive slope. (c) Calculate the break-even point by setting $y=0$.
choose m and b → equation → break-even at y=0 or set revenue = cost → solve →
$d=18-23=-5$ and $a_1-d=23-(-5)=28$, so $y=-5x+28$.
Problem 4b
At $x=1$, $y=-5+28=23$. At $x=3$, $y=-15+28=13$.
Problem 4c
The slope is $-5$, exactly equal to the signed common difference $d=-5$.
Problem 5a
$11=\frac32(4)+b=6+b$, so $b=5$.
Problem 5b
$y=\frac32x+5$.
Problem 6a
$\Delta y=13-5=8$.
Problem 6b
$\Delta x=6-2=4$.
Problem 6c
$m=\frac{8}{4}=2$ y-units per x-unit.
Problem 6d
$5=2(2)+b$, so $b=1$.
Problem 6e
$y=2x+1$.
Problem 6f
$y=2(10)+1=21$.
Problem 7a
$y_1=12x+50$.
Problem 7b
$y_2=5x+200$.
Problem 7c
$12x+50=5x+200$, so $x=150/7\approx21.4$ weeks (during week 22).
Problem 8a
Example: choose $m=3$. Then $20=3(4)+b$, so $b=8$ and $y=3x+8$.
Problem 8b
Example: choose $m=-1$. Then $20=-4+b$, so $b=24$ and $y=-x+24$.
Problem 8c
No. Equal slopes with different intercepts make parallel lines.
Problem 9a
Example: $x=$ riding lessons taught; $y=$ net profit in euros. Mia spends €90 on supplies and earns €15 per lesson.
Problem 9b
$y=15x-90$.
Problem 9c
$0=15x-90$, so break-even occurs at $x=6$ lessons.
What this unlocks next
You can now connect a context, an equation, and a graph. Future maths, science, and economics lessons will reuse this skill whenever one quantity changes steadily with another.