Module 14 of 47Lesson 1 of 8 · Functions 1 · Gradient, intercepts & y = mx + c
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Straight lines & functions · Essential understanding 1

Gradient, intercepts & y = mx + c

IB AI SL 2.1with Prof. Kitso
💻 Using the TI-Nspire CX Student Software on a laptop
  • The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
  • menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
  • enter = Enter  ·  tab = Tab (next box)  ·  esc = Esc  ·  del = Backspace.
  • Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
  • Fraction: type /  ·  √: type sqrt(  ·  π: type pi  ·  eˣ: type exp(  ·  store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
  • New page: Ctrl+I (or the Insert menu)  ·  change page: Ctrl+← / Ctrl+→  ·  graph entry line: Ctrl+G  ·  var: Ctrl+L.
  • Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
  • IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
  • If you type nDeriv(, the software shows it as centralDiff( — it is the same command.
The big idea: A straight line is described by two numbers: its gradient (how steep it is) and its y-intercept (where it crosses the y-axis). Together they give the equation y = mx + c.
By the end of this lesson you can:
  • calculate the gradient of a line from two points
  • read the gradient and y-intercept from y = mx + c
  • find the x-intercept and y-intercept of a line
  • write down the equation of a line from its gradient and a point, or from two points
  • explain what the gradient and intercept mean in a real situation
🧑🏾 Imagine this…
You are walking up a road near Gaborone. After every 10 m forward, the road rises 1 m. Another road rises 3 m for every 10 m forward. Which is steeper? The second one, of course. The gradient is just that comparison written as a number: how much up, for each step across.
Gradient (m)
how steep a line is. (“rise over run”). It is also the change in y when x goes up by 1.
y-intercept (c)
the y-value where the line crosses the y-axis. This is where x = 0, so the point is (0, c).
x-intercept
the x-value where the line crosses the x-axis. This is where y = 0, so the point is (x, 0).
Linear function
a rule whose graph is a straight line, for example f(x) = 2x + 1.
★ The rules
  • Gradient from two points and :     (in the formula booklet)
  • Gradient-intercept form: y = mx + c, where m is the gradient and c is the y-intercept.
  • m > 0: line goes up to the right. m < 0: line goes down to the right. m = 0: horizontal line y = k.
  • A vertical line x = h has an undefined gradient (we cannot divide by zero).
  • To find the y-intercept: put x = 0. To find the x-intercept: put y = 0 and solve.
⚠️ Mistakes that lose marks
  • Subtract the y-values and the x-values in the same order. (y₂ − y₁) over (x₂ − x₁) is right; (y₂ − y₁) over (x₁ − x₂) gives the wrong sign.
  • The line must be written as y = … before you read off m. In 4y = 2x + 10 the gradient is not 2: divide everything by 4 first.
  • Brackets on the calculator: type (9-1)/(6-2), not 9-1/6-2.
💡 Check your sign by looking
Sketch the two points roughly. If the line goes uphill to the right, your gradient must be positive. If you got a negative number, go back and check.
🧪 Try it yourselfPlay with m and c
Drag m and c. Set m = 2 and c = 1, then change m to −2. What happens to the line? Where does it cut the axes when m = 2 and c = −4?

Worked examples

Example 1.1Gradient from two points
Find the gradient of the line through each pair of points.
  1. (a) (2, 1) and (6, 9)
  2. (b) (−1, 7) and (3, −1)
  3. (c) (−3, 4) and (5, 4)
  4. (d) (2, −1) and (2, 5)
🧠 Think it through
  1. Use each time. Always keep the same order top and bottom.
  2. Part (c): both y-values are the same, so there is no rise: the line is flat.
  3. Part (d): both x-values are the same, so the line is vertical.
✍️ By hand
(a)
(b)
(c)
(d) → undefined
🔢 On your TI-Nspire CX II
  1. Open a Calculator page: ⌂ on💻 click ⌂ on keypad 1. Always put brackets round the top and the bottom.
  2. Type (9-1)/(6-2) enter ↵💻 Enter, then (-1-7)/(3-(-1)) enter ↵💻 Enter. Use the grey (-) key for a negative sign, not the minus key.
    1.1*DocRAD9−16−22-1−73−(-1)-24−45−(-3)0
    For part (d) the calculator gives a “Divide by zero” error: the sign of a vertical line
✅ Answer (a) m = 2   (b) m = −2   (c) m = 0 (horizontal line y = 4)   (d) undefined (vertical line x = 2).
On exam papers a vertical line is written x = 2. Do not write “gradient = 0” for a vertical line: zero is for flat lines.
Example 1.2Reading m and c, and finding the intercepts
Write down the gradient and the y-intercept of each line, then find where it crosses the x-axis.
  1. (a) y = 3 − 2x
  2. (b) 4y = 2x + 10
🧠 Think it through
  1. Rewrite in the form y = mx + c. In (a) just reorder. In (b) divide every term by 4.
  2. The number in front of x is m. The number on its own is c.
  3. For the x-intercept put y = 0 and solve for x.
✍️ By hand
(a) y = −2x + 3, so and . Put y = 0: →
(b) , so and . Put y = 0: →
🔢 On your TI-Nspire CX II
  1. To check the x-intercept, open a Graphs page: ctrl💻 Ctrl doc ▾💻 File menu then 2 (Add Graphs).
  2. Type 3-2x next to f1(x)= and press enter ↵💻 Enter. Then go to menu💻 click menu on keypad▸6: Analyze Graph▸1: Zero. Click to the left of the crossing, press enter ↵💻 Enter, click to the right, press enter ↵💻 Enter.
    1.1*DocRAD-366-6xy22f1(x)=3−2·xZeroX=1.5
✅ Answer (a) m = −2, c = 3, x-intercept 1.5, so it crosses at (1.5, 0). (b) m = 0.5, c = 2.5, x-intercept −5, so it crosses at (−5, 0).
Example 1.3Gradient in a real situation
A water tank at a lodge in Kasane is being filled. After 2 minutes it holds 40 litres. After 6 minutes it holds 100 litres. The volume V (litres) after t minutes is a straight-line graph.
  1. (a) Find the gradient and say what it means.
  2. (b) Find the equation of V in terms of t.
  3. (c) How long until the tank holds 250 litres?
🧠 Think it through
  1. The two points on the line are (2, 40) and (6, 100).
  2. . The units are litres per minute.
  3. Use V = mt + c. Put in one point to find c. The number c is the volume at t = 0.
  4. For (c) put V = 250 and solve for t.
✍️ By hand
(a)
(b) . Use (2, 40): → . So .
(c) → →
🔢 On your TI-Nspire CX II
  1. On a Calculator page type (100-40)/(6-2) enter ↵💻 Enter.
  2. Solve (c) with menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve: type nSolve(15t+10=250,t) enter ↵💻 Enter.
    1.1*DocRAD100−406−215nSolve(15·t+10=250,t)16.
✅ Answer (a) m = 15: the tank gains 15 litres every minute. (b) V = 15t + 10. (c) 16 minutes.
c = 10 means there were already 10 litres in the tank when we started the clock (t = 0). Always say what c means in context.
Example 1.4Equation of a line: a point and a gradient, or two points
Find the equation of the line (a) with gradient 3 that passes through (2, 5); (b) that passes through (1, 4) and (5, 12).
🧠 Think it through
  1. Write y = mx + c. You need m and c.
  2. (a) m is given. Put the point in to find c.
  3. (b) First find m from the two points. Then use either point to find c. Check with the other point.
✍️ By hand
(a) . Point (2, 5): → . So .
(b) . Point (1, 4): → . So .
Check with (5, 12): ✓
🔢 On your TI-Nspire CX II
  1. Use the calculator only for the arithmetic. Type (12-4)/(5-1) enter ↵💻 Enter to get m = 2.
  2. Check the point (5, 12): type 2*5+2 enter ↵💻 Enter.
    1.1*DocRAD12−45−122·5+212
✅ Answer (a) y = 3x − 1   (b) y = 2x + 2
Always substitute the second point as a check. It takes five seconds and finds sign errors.
Example 1.5Exam style: gradient, equation and intercept
The line L passes through A(−2, 5) and B(4, −4).
  1. (a) Find the gradient of L.
  2. (b) Find the equation of L in the form y = mx + c.
  3. (c) Find the x-intercept of L.
  4. (d) Show that the point (2, −1) lies on L.
🧠 Think it through
  1. Find means you show your working and get a number. Start with the gradient formula.
  2. Use A to find c, then check with B.
  3. Show that means you must substitute and show both sides are equal. Do not just say “yes”.
✍️ By hand
(a)
(b) . Point A: → . So .
(c) Put y = 0: →
(d) When x = 2: , which is the y-value of the point. ✓
🔢 On your TI-Nspire CX II
  1. Graphs page: type -1.5x+2 in f1(x). Use menu💻 click menu on keypad▸6: Analyze Graph▸1: Zero to see the x-intercept.
  2. Or use the Calculator page: nSolve(-1.5x+2=0,x) enter ↵💻 Enter.
    📍 Where to find it on the TI-Nspire
    nSolve( is in menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve
    1.1*DocRAD-4−54−(-2)-1.5nSolve(-1.5·x+2=0,x)1.33333-1.5·2+2-1
    1.11.2*DocRAD-468-6xy22f1(x)=-1.5·x+2ZeroX=1.33333
✅ Answer (a) m = −1.5   (b) y = −1.5x + 2   (c) x = 1.33 (3 s.f.)   (d) at x = 2, y = −1, so (2, −1) is on the line.
The exact answer to (c) is . Give 1.33 as a 3 s.f. answer, or leave it as the fraction.

Challenge questions

⭐ Challenge 1.6Line through P and QPaper 1 style
The line through P(3, −2) and Q(7, 10) crosses the x-axis at R.
  1. (a) Find the gradient of PQ.
  2. (b) Find the equation of PQ.
  3. (c) Find the coordinates of R.
✏️ Try it on paper and on your calculator first. Then check.
Show the full solution
🧠 Think it through
  1. Gradient from P and Q.
  2. Use P to find c, check with Q.
  3. R is on the x-axis, so its y-coordinate is 0.
✍️ By hand
(a)
(b) → , so . Check Q: ✓
(c) →
🔢 On your TI-Nspire CX II
  1. Type (10-(-2))/(7-3) enter ↵💻 Enter, then nSolve(3x-11=0,x) enter ↵💻 Enter.
    📍 Where to find it on the TI-Nspire
    nSolve( is in menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve
1.1*DocRAD10−(-2)7−33nSolve(3·x−11=0,x)3.66667
✅ Answer (a) 3   (b) y = 3x − 11   (c) R = (3.67, 0), exact .
⭐ Challenge 1.7Camping at the OkavangoPaper 2 style
A camp in the Okavango Delta charges a fixed booking fee plus a fee for each night. A group paid P680 for 2 nights and P1 400 for 5 nights. The cost C (pula) for n nights is a linear function of n.
  1. (a) Find the gradient of the graph of C against n and say what it represents.
  2. (b) Find the equation of C in terms of n.
  3. (c) State what the y-intercept represents.
  4. (d) Find the cost for 8 nights.
  5. (e) Kabo has P2 000. What is the greatest number of whole nights she can pay for?
✏️ Try it on paper and on your calculator first. Then check.
Show the full solution
🧠 Think it through
  1. Two points on the line: (2, 680) and (5, 1400).
  2. The gradient is the cost per night. The intercept is the fixed fee.
  3. For (e) solve C ≤ 2000, then round DOWN, because you cannot book part of a night.
✍️ By hand
(a) . It is the cost of one extra night (P240 per night).
(b) ; → . So .
(c) P200 is the fixed booking fee (the cost when n = 0).
(d)
(e) → , so 7 nights.
🔢 On your TI-Nspire CX II
  1. Calculator page: (1400-680)/(5-2) enter ↵💻 Enter, then 240*8+200 enter ↵💻 Enter.
  2. For (e) solve the equal case: nSolve(240n+200=2000,n) gives 7.5. Round down to 7.
    📍 Where to find it on the TI-Nspire
    nSolve( is in menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve
1.1*DocRAD1400−6805−2240240·8+2002120nSolve(240·n+200=2000,n)7.5
✅ Answer (a) 240: P240 per night. (b) C = 240n + 200. (c) The fixed booking fee, P200. (d) P2 120. (e) 7 nights (7.5 rounded down; 7 nights cost P1 880 and 8 nights cost P2 120).
Next: Lesson 2 · Forms of the equation of a line
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