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Graphs of functions

E2.10 Graphs of Functions

1. Functions You Must Know

ax^n where n = -2, -1, -1/2, 0, 1/2, 1, 2, 3

ab^x + c where b is positive integer

Sums of up to 3 terms

2. How to Draw Any Graph

1. Make table with x = -3, -2, -1, 0, 1, 2, 3 (skip undefined)

2. Calculate y for each x

3. Plot points accurately

4. Draw smooth curve through points

Example 1: y = 2x + 3/x²

x-3-2-1123
y-5.67-3.25154.756.33

Two curves, asymptote at x=0

Example 2: y = ¼ × 2ˣ

x-2-10123
y0.06250.1250.250.512

Exponential growth, asymptote y=0

3. Solving Equations Graphically

Root: Where graph crosses x-axis

Intersection: Where two graphs cross

Example: Solve x³ + x - 4 = 0

Plot y = x³ + x - 4

At x=1, y=-2

At x=2, y=6

Root between 1 and 2 → x ≈ 1.38

4. Exponential Graphs

b > 1 → Growth (e.g., population)

0 < b < 1 → Decay (e.g., radioactive)

y = ab^x + c has asymptote y = c

5. Exam Tips

  • Label x and y axes
  • Use sharp pencil
  • Ruler for straight lines only
  • Show asymptotes with dashed lines
  • For 1/x, don't connect points across x=0

6. Common Graph Shapes

x² → Parabola

x³ → S-shape

1/x → Two hyperbola curves

√x → Half parabola (x≥0)

2ˣ → Exponential curve

E2.10 Graphs of Functions - Complete Cambridge IGCSE Notes

1. Syllabus Coverage

E2.10: Interpret and use graphs in practical situations including travel graphs and conversion graphs.

E2.11: Construct tables of values and draw graphs for functions of the form axn and abx + c. Solve associated equations graphically.

E2.12: Estimate gradients of curves by drawing tangents.

E2.13: Understand the idea of a derived function. Use derivatives of axn functions.

2. Linear Graphs: y = mx + c

Gradient m = (y₂ - y₁)/(x₂ - x₁)

Example: Find equation through (1,3) and (3,7)

Gradient m = (7-3)/(3-1) = 2

Equation: y = 2x + c

Substitute (1,3): 3 = 2(1) + c → c = 1

Answer: y = 2x + 1

3. Plotting Curves: Method

For y = 2x² + x - 6, -3 ≤ x ≤ 3:

x-3-2-10123
y90-5-6-3415

Steps: 1. Complete table 2. Plot points 3. Draw smooth curve

4. Exponential Functions: y = abˣ + c

Example: Bacteria growth y = 4 × 2x-1

At start (x=0): y = 4 × 2-1 = 2 bacteria

At x=3: y = 4 × 22 = 16 bacteria

5. Gradient of Curves

Draw tangent to curve at point, find gradient of tangent

Gradient = (vertical change)/(horizontal change)

6. Solving Equations Graphically

Example: Solve 2x² - x - 3 = 6

1. Draw y = 2x² - x - 3

2. Draw horizontal line y = 6

3. Solutions: x-coordinates of intersections: x ≈ -1.9 and x ≈ 2.4

7. Distance-Time Graphs

Gradient = speed

Flat section: stationary (speed = 0)

Straight line: constant speed

Curved: acceleration/deceleration

8. Speed-Time Graphs

Gradient = acceleration

Area under graph = distance

Distance = area of triangle + area of rectangle

9. Differentiation (E2.13)

If y = xn, then dy/dx = nxn-1

Example: y = x² - 4x + 1

dy/dx = 2x - 4

Gradient at x=3: 2(3) - 4 = 2

10. Turning Points

Set dy/dx = 0 to find turning points

Example: f(x) = 3x³ - 2x²

f'(x) = 9x² - 4x

Set 9x² - 4x = 0 → x(9x-4)=0

Turning points: (0,0) and (4/9, -32/243)

11. Exam Questions - What They Ask

Type 1: Complete Table & Plot

"Complete table for y = x² - 1/x, then plot graph"

Type 2: Solve Graphically

"Use graph to solve x² - 1/x = -3x"

Type 3: Gradient/Tangent

"Draw tangent at x=-2, estimate gradient"

Type 4: Real-world Graphs

"Bacteria: N = 1000 × 1.4ˣ. Draw graph, find when N=3000"

Type 5: Speed-Time/Distance-Time

"Calculate acceleration, distance, average speed"

12. Key Formulas to Memorize

1. Gradient of line through (x₁,y₁) and (x₂,y₂): m = (y₂-y₁)/(x₂-x₁)
2. Equation of line: y = mx + c
3. Differentiation: d/dx(xn) = nxn-1
4. Distance from speed-time: Area under graph

13. Common Mistakes

  • Joining points with straight lines on curves
  • Forgetting asymptotes for 1/x functions
  • Incorrect gradient calculation: y/x instead of Δy/Δx
  • Not showing working for differentiation
  • Inaccurate plotting from tables

14. Exam Technique

1. Use sharp pencil for graphs

2. Label axes clearly

3. Show tangent construction lines

4. Give answers to required accuracy (1 decimal place, etc.)

5. Show all steps in differentiation

6. Check domain: x ≠ 0 for 1/x functions

15. Past Paper Focus

June 2007 Q18: Equation of parallel line

June 2008 Q9: Exponential growth N = 1000 × 1.4ˣ

November 2008 Q16: Using graph to solve f(x) = k

November 2008 Q3: Complete table, plot, find gradient at point

June 2006 Q1: Speed-time graph calculations

16. Essential Practice

1. Plot y = x² - 1/x for -3 ≤ x ≤ 3 (exclude x=0)

2. Solve x² - 2x - 3 = 0 graphically

3. Find gradient of y = x³ - 2x² at x=2 using differentiation

4. Calculate distance from speed-time graph with areas

5. Find equation of tangent to curve at given point

Vertex Form Notes - Linear and Quadratic Equations

VERTEX FORM NOTES

Linear and Quadratic Equations

1. LINEAR EQUATIONS - STANDARD FORMS

Gradient-Intercept Form (y = mx + c)

Form: y = mx + c

Where:

  • m = gradient (slope)
  • c = y-intercept (where line crosses y-axis)
  • x and y are coordinates

Example: y = 3x + 2

  • Gradient = 3
  • y-intercept = 2
  • Line goes through (0, 2) and rises 3 units for every 1 unit right

Point-Gradient Form

Form: y - y₁ = m(x - x₁)

Where:

  • m = gradient
  • (x₁, y₁) = a point on the line

Use: When you know gradient and one point

Example: Line through (2, 5) with gradient 3

y - 5 = 3(x - 2) y - 5 = 3x - 6 y = 3x - 1

General Form

Form: ax + by + c = 0

Where a, b, c are constants

Example: 2x + 3y - 6 = 0

2. QUADRATIC EQUATIONS - VERTEX FORM

What is Vertex Form?

y = a(x - h)² + k

Where:

  • (h, k) = coordinates of the VERTEX (turning point)
  • a = coefficient that affects:
    • Shape (width of parabola)
    • Direction (opens up if a > 0, down if a < 0)

Key Features

The Vertex:

  • Minimum point if a > 0 (parabola opens upward ∪)
  • Maximum point if a < 0 (parabola opens downward ∩)
  • Located at (h, k)

The value of 'a':

  • |a| > 1: Narrow parabola (stretched vertically)
  • |a| < 1: Wide parabola (compressed vertically)
  • a > 0: Opens upward (smiles ☺)
  • a < 0: Opens downward (frowns ☹)

Reading Vertex Form

Example 1: y = (x - 3)² + 2

Compare with: y = a(x - h)² + k a = 1 h = 3 k = 2 Vertex = (3, 2) Opens upward (a = 1 > 0) Minimum value = 2 at x = 3

Example 2: y = -2(x + 1)² + 5

Rewrite as: y = -2(x - (-1))² + 5 a = -2 h = -1 k = 5 Vertex = (-1, 5) Opens downward (a = -2 < 0) Maximum value = 5 at x = -1 Narrower than normal (|a| = 2 > 1)

Example 3: y = ½(x - 4)² - 3

a = ½ h = 4 k = -3 Vertex = (4, -3) Opens upward (a = ½ > 0) Minimum value = -3 at x = 4 Wider than normal (|a| = ½ < 1)

Turning Point Formula

For a quadratic in the form y = ax² + bx + c, the x-coordinate of the vertex is:

x = -b / 2a
  • a: coefficient of x²
  • b: coefficient of x
  • y: find by plugging x back into the original equation

3. CONVERTING BETWEEN FORMS

From Standard Form to Vertex Form

Standard form: y = ax² + bx + c

Method: Completing the Square

Step-by-step:

Example: Convert y = x² + 6x + 5 to vertex form

Step 1: Group x terms y = (x² + 6x) + 5 Step 2: Complete the square - Take half of coefficient of x: 6 ÷ 2 = 3 - Square it: 3² = 9 - Add and subtract inside bracket y = (x² + 6x + 9 - 9) + 5 y = (x² + 6x + 9) - 9 + 5 Step 3: Write as perfect square y = (x + 3)² - 4 Step 4: Rewrite in standard vertex form y = (x - (-3))² + (-4) Vertex = (-3, -4)

Example 2: y = 2x² - 12x + 10

Step 1: Factor out coefficient of x² y = 2(x² - 6x) + 10 Step 2: Complete the square inside bracket - Half of -6 = -3 - Square: (-3)² = 9 y = 2(x² - 6x + 9 - 9) + 10 y = 2(x² - 6x + 9) - 2(9) + 10 y = 2(x² - 6x + 9) - 18 + 10 Step 3: Simplify y = 2(x - 3)² - 8 Vertex = (3, -8)

From Vertex Form to Standard Form

Method: Expand the brackets

Example: y = (x - 2)² + 3

Step 1: Expand (x - 2)² (x - 2)² = (x - 2)(x - 2) = x² - 2x - 2x + 4 = x² - 4x + 4 Step 2: Add constant y = x² - 4x + 4 + 3 y = x² - 4x + 7

Example 2: y = -3(x + 1)² - 2

Step 1: Expand (x + 1)² (x + 1)² = x² + 2x + 1 Step 2: Multiply by -3 -3(x² + 2x + 1) = -3x² - 6x - 3 Step 3: Add constant y = -3x² - 6x - 3 - 2 y = -3x² - 6x - 5

4. USING VERTEX FORM

Finding Maximum/Minimum Values

For y = a(x - h)² + k:

If a > 0:

  • Minimum value = k
  • Occurs at x = h

If a < 0:

  • Maximum value = k
  • Occurs at x = h

Example 1: y = (x - 5)² + 1

a = 1 > 0, so minimum Minimum value = 1 Occurs when x = 5

Example 2: y = -2(x + 3)² + 7

a = -2 < 0, so maximum Maximum value = 7 Occurs when x = -3

Graphing from Vertex Form

Steps:

  1. Identify vertex (h, k)
  2. Plot vertex
  3. Determine if opens up or down (sign of a)
  4. Find additional points by substituting x-values
  5. Use symmetry about vertex

Example: y = (x - 1)² - 4

Step 1: Vertex = (1, -4) Step 2: Opens upward (a = 1 > 0) Step 3: Find additional points When x = 0: y = (0 - 1)² - 4 = 1 - 4 = -3 → (0, -3) When x = 2: y = (2 - 1)² - 4 = 1 - 4 = -3 → (2, -3) When x = 3: y = (3 - 1)² - 4 = 4 - 4 = 0 → (3, 0) When x = -1: y = (-1 - 1)² - 4 = 4 - 4 = 0 → (-1, 0) Step 4: Symmetry: Points are symmetric about x = 1

Finding x-intercepts

Set y = 0 and solve

Example: y = (x - 3)² - 9

0 = (x - 3)² - 9 9 = (x - 3)² ±3 = x - 3 x - 3 = 3 or x - 3 = -3 x = 6 or x = 0 x-intercepts: (0, 0) and (6, 0)

Finding y-intercept

Set x = 0 and solve

Example: y = 2(x - 1)² + 3

y = 2(0 - 1)² + 3 y = 2(1) + 3 y = 5 y-intercept: (0, 5)

5. WORKED EXAMPLES

Example 1: Complete the square

Question: Write y = x² + 8x + 11 in vertex form

Solution: y = x² + 8x + 11 y = (x² + 8x) + 11 Half of 8 = 4, square it = 16 y = (x² + 8x + 16 - 16) + 11 y = (x + 4)² - 16 + 11 y = (x + 4)² - 5 Vertex form: y = (x - (-4))² + (-5) Vertex: (-4, -5) Minimum value: -5

Example 2: With coefficient ≠ 1

Question: Write y = 3x² - 18x + 20 in vertex form

Solution: y = 3x² - 18x + 20 y = 3(x² - 6x) + 20 Half of -6 = -3, square it = 9 y = 3(x² - 6x + 9 - 9) + 20 y = 3(x² - 6x + 9) - 3(9) + 20 y = 3(x - 3)² - 27 + 20 y = 3(x - 3)² - 7 Vertex: (3, -7) Minimum value: -7

Example 3: Negative coefficient

Question: Write y = -x² + 4x - 1 in vertex form

Solution: y = -x² + 4x - 1 y = -(x² - 4x) - 1 Half of -4 = -2, square it = 4 y = -(x² - 4x + 4 - 4) - 1 y = -(x² - 4x + 4) - (-4) - 1 y = -(x - 2)² + 4 - 1 y = -(x - 2)² + 3 Vertex: (2, 3) Maximum value: 3

Example 4: Using vertex form

Question: A parabola has vertex at (2, -5) and passes through (0, -1). Find the equation.

Solution: Use: y = a(x - h)² + k Vertex (2, -5), so h = 2, k = -5 y = a(x - 2)² - 5 Point (0, -1) lies on curve: -1 = a(0 - 2)² - 5 -1 = 4a - 5 4 = 4a a = 1 Equation: y = (x - 2)² - 5

Example 5: Finding range

Question: Find the range of y = -(x + 1)² + 4

Solution: Vertex form: y = -(x - (-1))² + 4 Vertex: (-1, 4) a = -1 < 0, so opens downward Maximum value = 4 Range: y ≤ 4 or (-∞, 4]

6. QUICK REFERENCE

Vertex Form Template

y = a(x - h)² + k
  • Vertex: (h, k)
  • If a > 0: Opens up ∪, minimum at k
  • If a < 0: Opens down ∩, maximum at k
  • If |a| > 1: Narrow parabola
  • If |a| < 1: Wide parabola

Completing the Square Steps

For y = ax² + bx + c:

  1. Factor out 'a' from x terms: y = a(x² + (b/a)x) + c
  2. Take half of (b/a): (b/2a)
  3. Square it: (b/2a)²
  4. Add and subtract inside: a(x² + (b/a)x + (b/2a)² - (b/2a)²) + c
  5. Factor perfect square: a(x + b/2a)² - a(b/2a)² + c
  6. Simplify: y = a(x - h)² + k

Common Sign Mistakes to Avoid

Be careful with signs!

❌ y = (x + 3)² - 2 has vertex (3, -2) ✓ y = (x + 3)² - 2 has vertex (-3, -2) Remember: y = a(x - h)² + k - If you see (x + 3), that means (x - (-3)), so h = -3

More examples:

  • y = (x - 5)² + 1 → vertex is (5, 1)
  • y = (x + 2)² - 3 → vertex is (-2, -3)
  • y = -(x - 1)² + 4 → vertex is (1, 4)
  • y = 2(x + 4)² - 1 → vertex is (-4, -1)

7. PRACTICE PROBLEMS

Problem 1

Write in vertex form and find the vertex:

a) y = x² + 10x + 21

b) y = x² - 6x + 4

c) y = 2x² + 8x - 3

Problem 2

Expand to standard form:

a) y = (x - 4)² + 1

b) y = 3(x + 2)² - 5

c) y = -2(x - 1)² + 7

Problem 3

Find the maximum or minimum value:

a) y = (x - 2)² - 7

b) y = -(x + 3)² + 5

c) y = 2(x - 1)² + 3

SOLUTIONS TO PRACTICE PROBLEMS

Problem 1 Solutions

a) y = x² + 10x + 21

y = (x² + 10x) + 21 y = (x² + 10x + 25 - 25) + 21 y = (x + 5)² - 25 + 21 y = (x + 5)² - 4 Vertex: (-5, -4)

b) y = x² - 6x + 4

y = (x² - 6x) + 4 y = (x² - 6x + 9 - 9) + 4 y = (x - 3)² - 9 + 4 y = (x - 3)² - 5 Vertex: (3, -5)

c) y = 2x² + 8x - 3

y = 2(x² + 4x) - 3 y = 2(x² + 4x + 4 - 4) - 3 y = 2(x + 2)² - 8 - 3 y = 2(x + 2)² - 11 Vertex: (-2, -11)

Problem 2 Solutions

a) y = (x - 4)² + 1

y = (x² - 8x + 16) + 1 y = x² - 8x + 17

b) y = 3(x + 2)² - 5

y = 3(x² + 4x + 4) - 5 y = 3x² + 12x + 12 - 5 y = 3x² + 12x + 7

c) y = -2(x - 1)² + 7

y = -2(x² - 2x + 1) + 7 y = -2x² + 4x - 2 + 7 y = -2x² + 4x + 5

Problem 3 Solutions

a) y = (x - 2)² - 7
a = 1 > 0, opens upward
Minimum value = -7 at x = 2

b) y = -(x + 3)² + 5
a = -1 < 0, opens downward
Maximum value = 5 at x = -3

c) y = 2(x - 1)² + 3
a = 2 > 0, opens upward
Minimum value = 3 at x = 1

KEY TAKEAWAYS

  • ✓ Vertex form makes it easy to identify turning point
  • ✓ Vertex (h, k) is the maximum or minimum point
  • ✓ Sign of a tells you if parabola opens up or down
  • ✓ Completing the square converts standard to vertex form
  • ✓ Watch signs carefully when identifying h from (x - h)
  • ✓ Vertex form is useful for graphing and finding range

End of Vertex Form Notes

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