Quadratic Graphs – Grade 9 IGCSE Maths

a > 0 → U-shape

Positive coefficient gives a minimum turning point

x = −b/(2a)

Line of symmetry and x-coordinate of vertex

y-intercept = c

Set x = 0 to find where graph crosses the y-axis

ax²+bx+c = 0

Solve for roots — where the graph crosses the x-axis

Shape & Orientation

U-shape vs n-shape based on sign of a

Turning Point

Vertex using x = −b/(2a) formula

Roots & Intercepts

Solving ax²+bx+c = 0 for x-intercepts

Transformations

Shifts, stretches and reflections of parabolas

Parabola Plotter

Adjust a, b, c with sliders and see the graph live

Practice 20q

Test all quadratic graph skills

1. The Standard Form y = ax² + bx + c

Every quadratic graph is a parabola. The three constants a, b and c each control a different feature of the curve.

y = ax² + bx + c (a ≠ 0)

a controls the shape and width: large |a| = narrow, small |a| = wide
b shifts the axis of symmetry left or right
c is the y-intercept — the value of y when x = 0

Always identify a, b and c first before attempting to sketch or analyse a quadratic.

2. Shape: U-shape or ∩-shape?

a > 0 → U-shape (opens upward) → has a minimum turning point
a < 0 → ∩-shape (opens downward) → has a maximum turning point

Remember: positive a = happy parabola (smiling U). Negative a = sad parabola (frowning ∩).

3. Vertex / Turning Point

The vertex (turning point) is the minimum or maximum of the parabola.

x-coordinate of vertex: x = −b / (2a)

Substitute this x value back into y = ax² + bx + c to get the y-coordinate.

Example: y = x² − 4x + 3, so a=1, b=−4, c=3
x = −(−4)/(2×1) = 4/2 = 2
y = (2)² − 4(2) + 3 = 4 − 8 + 3 = −1
Vertex = (2, −1)

The line of symmetry passes through the vertex: equation is x = −b/(2a).

4. y-intercept

Set x = 0: y = c

The y-intercept is always the constant term c in y = ax² + bx + c.

y = 2x² + 4x − 6: y-intercept is (0, −6)
y = x² − 2x + 5: y-intercept is (0, 5)

5. x-intercepts (Roots)

Where the parabola crosses the x-axis. Set y = 0 and solve ax² + bx + c = 0 by factorising, completing the square, or the quadratic formula.

x = (−b ± √(b²−4ac)) / (2a)

y = x² − 4x + 3: factorise to (x−1)(x−3) = 0 → roots x = 1 and x = 3
y = x² − 4: x² = 4 → x = ±2
No real roots if discriminant b² − 4ac < 0 (graph doesn't cross x-axis)

Number of roots from discriminant:
b²−4ac > 0 → two distinct roots
b²−4ac = 0 → one repeated root (touches x-axis)
b²−4ac < 0 → no real roots

6. Transformations of y = f(x)

y = f(x) + a — vertical shift: moves graph up by a (down if a negative)
y = f(x + a) — horizontal shift: moves graph left by a (right if a negative)
y = af(x) — vertical stretch: scale factor a from the x-axis
y = −f(x) — reflection in the x-axis (flips U to ∩)

For y = f(x + a): the + inside the bracket moves left — the opposite direction to what you expect!

Examples from y = x²:
y = x² + 3 → shift up 3
y = (x − 2)² → shift right 2
y = 2x² → vertical stretch ×2 (narrower)
y = −x² → reflection in x-axis

7. Reading from Graphs

From a sketch you can read:
• Roots: x-coordinates where graph crosses/touches x-axis
• y-intercept: where graph crosses y-axis
• Turning point: coordinates of minimum or maximum
• Line of symmetry: vertical line through vertex
• Whether a is positive or negative (U vs ∩ shape)

To find gradient at a point on a parabola, draw a tangent to the curve at that point and calculate rise/run of the tangent.

Example 1 — y = x² − 4x + 3: full analysis

Identify: a=1, b=−4, c=3. Since a>0, U-shaped with a minimum.

Vertex: x = −(−4)/(2×1) = 2. y = 4 − 8 + 3 = −1. Turning point: (2, −1)

y-intercept: x=0 → y = 3. Point: (0, 3)

Roots: x²−4x+3 = 0 → (x−1)(x−3) = 0 → x = 1 and x = 3

Line of symmetry: x = 2

Example 2 — y = −x² + 2x + 3: maximum parabola

Identify: a=−1, b=2, c=3. Since a<0, ∩-shaped with a maximum.

Vertex: x = −2/(2×−1) = −2/(−2) = 1. y = −1 + 2 + 3 = 4. Turning point: (1, 4)

y-intercept: y = 3. Point: (0, 3)

Roots: −x²+2x+3 = 0 → x²−2x−3 = 0 → (x−3)(x+1) = 0 → x = 3 and x = −1

Example 3 — y = 2x² + 4x − 6

Identify: a=2, b=4, c=−6. U-shaped minimum.

Vertex: x = −4/(2×2) = −1. y = 2(1) + 4(−1) − 6 = 2 − 4 − 6 = −8. Turning point: (−1, −8)

y-intercept: y = −6. Point: (0, −6)

Roots: 2x²+4x−6 = 0 → x²+2x−3 = 0 → (x+3)(x−1) = 0 → x = −3 and x = 1

Example 4 — Sketch y = x² − 2x − 8 showing all key features

Shape: a=1>0, so U-shaped minimum.

y-intercept: c = −8, so (0, −8)

Roots: x²−2x−8 = 0 → (x−4)(x+2) = 0 → x = 4 and x = −2

Vertex: x = −(−2)/(2) = 1. y = 1 − 2 − 8 = −9. Turning point: (1, −9)

Line of symmetry: x = 1 (midpoint of roots: (4 + (−2))/2 = 1 ✓)

Example 5 — Transformation: describe y = (x + 3)² − 5

Compare to y = x²:

y = f(x + 3) → translation 3 units left

y = f(x) − 5 → translation 5 units down

Combined: translation by vector (−3, −5). Vertex at (−3, −5).

Example 6 — Intersection of y = x² − 2x and y = x + 4

Set equal: x² − 2x = x + 4

x² − 3x − 4 = 0

(x − 4)(x + 1) = 0 → x = 4 or x = −1

Intersection points: when x=4, y=8 → (4, 8); when x=−1, y=3 → (−1, 3)

Parabola Plotter — y = ax² + bx + c

Adjust the sliders to change the parabola. The vertex is marked with a pink dot and roots with blue dots.

a: 1

b: -4

c: 3

Adjust sliders to see graph analysis.

Exercise 1 — Identify Key Features

For each equation, find the vertex (turning point), roots, and y-intercept. Enter the x-coordinate of the vertex and the y-intercept value.

Q1. y = x² + 2x − 3
x-coordinate of vertex: y-intercept:

Q2. y = x² − 6x + 5
x-coordinate of vertex: y-intercept:

Q3. y = −x² + 4 (b = 0)
x-coordinate of vertex: y-intercept:

Q4. y = x² − 4x + 4
x-coordinate of vertex: y-intercept:

Q5. y = 2x² − 8 (b = 0)
x-coordinate of vertex: y-intercept:

Exercise 2 — Table of Values

Calculate the y-value for y = x² − 3x + 2 at each given x value.

y = x² − 3x + 2. Substitute each x value carefully — watch the signs!

Q1. x = −1: y = (−1)² − 3(−1) + 2 = 1 + 3 + 2 = ?

Q2. x = 0: y = 0 − 0 + 2 = ?

Q3. x = 1: y = 1 − 3 + 2 = ?

Q4. x = 2: y = 4 − 6 + 2 = ?

Q5. x = 3: y = 9 − 9 + 2 = ?

Q6. x = 4: y = 16 − 12 + 2 = ?

Q7. For y = x² − 3x + 2, what is the y-intercept?

Q8. For y = x² − 3x + 2, what is the x-coordinate of the vertex? (Use x = −b/(2a))

Exercise 3 — Finding Turning Points

Use x = −b/(2a) to find the x-coordinate, then substitute to find y. Enter just the numerical values.

Q1. y = x² − 2x + 5. Vertex x-coordinate:

Q2. y = x² − 2x + 5. Vertex y-coordinate:

Q3. y = x² + 6x + 10. Vertex x-coordinate:

Q4. y = x² + 6x + 10. Vertex y-coordinate:

Q5. y = −x² + 4x − 1. Vertex x-coordinate:

Q6. y = −x² + 4x − 1. Vertex y-coordinate:

Q7. y = 2x² − 12x + 7. Vertex x-coordinate:

Q8. y = 2x² − 12x + 7. Vertex y-coordinate:

Exercise 4 — Transformations

Each equation is derived from y = x² by a single transformation. Enter the number for the transformation type: 1=shift up, 2=shift down, 3=shift right, 4=shift left, 5=vertical stretch, 6=reflection in x-axis.

Q1. y = x² + 5 — transformation number: by how much:

Q2. y = (x − 4)² — transformation number: by how much:

Q3. y = 3x² — transformation number: scale factor:

Q4. y = −x² — transformation number:

Q5. y = (x + 2)² — transformation number: by how much:

Q6. y = x² − 7 — transformation number: by how much:

Exercise 5 — Parabola and Line Intersections

Set the equations equal and solve the resulting quadratic. Enter the x-coordinates of intersection.

Q1. y = x² − 2x and y = x + 4. Smaller x-coordinate: Larger x-coordinate:

Q2. y = x² and y = x + 6. Smaller x: Larger x:

Q3. y = x² + 2x and y = 3x + 4. Smaller x: Larger x:

Q4. y = x² − 5x + 6 and y = 0. Roots at x = and x =

Q5. y = 2x² − 4 and y = x + 2. Smaller x: Larger x:

Practice — 20 Questions

Mixed questions on all quadratic graph topics. Enter numerical answers only.

P1. y = x² − 8x + 7: x-coordinate of vertex?

P2. y = x² − 8x + 7: y-intercept?

P3. y = x² − 8x + 7: y-coordinate of vertex?

P4. y = x² − 8x + 7: smaller root (x-intercept)?

P5. y = x² − 8x + 7: larger root?

P6. y = −2x² + 8x: x-coordinate of maximum?

P7. y = −2x² + 8x: maximum y-value?

P8. y = x² − 9: y-intercept?

P9. y = x² − 9: positive root?

P10. y = x² − 5x + 4: sum of the two roots?

P11. y = 3x² − 12x + 9: x-coordinate of vertex?

P12. y = 3x² − 12x + 9: y-intercept?

P13. y = x² + 4x − 5: smaller root?

P14. y = x² + 4x − 5: vertex y-coordinate?

P15. y = (x − 3)² + 2: x-coordinate of vertex?

P16. y = (x − 3)² + 2: minimum y-value?

P17. y = x² − 4x. At x = 5, what is y?

P18. y = 2x² − 5x − 3: y-intercept?

P19. y = x² − 2x − 15: larger root?

P20. y = −x² + 6x − 5: y-coordinate of maximum?

Challenge — 8 Harder Questions

Transformations, intersections, and reasoning. Some require multiple steps.

C1. The graph of y = x² is translated so the vertex is at (3, −4). Write the equation in the form y = (x − a)² + b. What is a?

C2. Same translation as C1. What is b?

C3. y = x² − 6x + k has a minimum value of −4. Find k.

C4. The parabola y = ax² + 4 passes through (2, 12). Find a.

C5. y = x² − 4x + 3 and y = −x + 3 intersect at two points. Find the smaller x-coordinate.

C6. Same intersection as C5. Find the larger x-coordinate.

C7. y = 2(x − 1)² − 8. What are the roots? Enter the positive root.

C8. A quadratic has roots at x = −2 and x = 5, and passes through (0, −10). Find the coefficient a in y = a(x+2)(x−5).