What is the general equation of a quadratic graph?

Quadratic graphs have the form y = ax² + bx + c and produce a parabola. Positive a gives an upright U-shape (∪); negative a gives an inverted U (∩). Key features include axes intercepts, roots, and the turning point (vertex). You need to sketch graphs, find intercepts, factorise for roots, and locate the turning point.

1. What are Quadratic Graphs?

The general equation of a quadratic graph is:

$y = ax^2 + bx + c$

The shape is called a parabola ("U" shape):

Positive quadratics ((a > 0)): parabola is upright ∪ (minimum turning point)
Negative quadratics ((a < 0)): parabola is upside down ∩ (maximum turning point)

2. Key Features of Quadratic Graphs

You need to be able to:

Sketch a quadratic graph from its equation or given information
Find axes intercepts:
- y-intercept: set (x = 0) → (y = c)
- x-intercepts (roots): set (y = 0) → solve (ax^2 + bx + c = 0)
Factorise (if possible) to find the roots
Find the turning point (maximum or minimum)

You may need to rearrange the equation into (y = ax^2 + bx + c) first.

3. Finding the Turning Point

The turning point (vertex) can be found by:

Completing the square: (y = a(x + p)^2 + q) gives turning point ((-p, q))
Symmetry: The x-coordinate of the vertex is midway between the two roots: (x = \frac{-b}{2a})
Calculator: Many calculators can find the turning point directly

For (a > 0): minimum at the vertex. For (a < 0): maximum at the vertex.

4. Worked Example: Sketching y = x² − 4x + 3

5. Worked Example: Negative Quadratic

6. Examiner Tips

Quadratic Graphs

Q: What shape does a negative quadratic (a < 0) have?

Inverted U-shape ∩ with a maximum turning point.

Q: How do you find the y-intercept of y = ax² + bx + c?

Set x = 0. The y-intercept is (0, c).

Q: How do you find the x-intercepts (roots)?

Set y = 0 and solve ax² + bx + c = 0 by factorising, completing the square, or the quadratic formula.

Q: What is the x-coordinate of the vertex of y = ax² + bx + c?

x = −b/(2a). It is the axis of symmetry.

Q: For y = x² − 4x + 4, how many x-intercepts?

One. x² − 4x + 4 = (x − 2)² = 0 gives x = 2 (repeated root). The parabola touches the x-axis at (2, 0).

Q: What is the turning point of y = (x − 3)² + 5?

(3, 5). In the form a(x + p)² + q, the vertex is (−p, q). Here (x − 3)² = (x + (−3))², so vertex (3, 5).

Summary

1. What are Quadratic Graphs?

The general equation of a quadratic graph is:

$y = ax^2 + bx + c$

The shape is called a parabola ("U" shape):

Positive quadratics ((a > 0)): parabola is upright ∪ (minimum turning point)
Negative quadratics ((a < 0)): parabola is upside down ∩ (maximum turning point)

2. Key Features of Quadratic Graphs

You need to be able to:

Sketch a quadratic graph from its equation or given information
Find axes intercepts:
- y-intercept: set (x = 0) → (y = c)
- x-intercepts (roots): set (y = 0) → solve (ax^2 + bx + c = 0)
Factorise (if possible) to find the roots
Find the turning point (maximum or minimum)

You may need to rearrange the equation into (y = ax^2 + bx + c) first.

3. Finding the Turning Point

The turning point (vertex) can be found by:

Completing the square: (y = a(x + p)^2 + q) gives turning point ((-p, q))
Symmetry: The x-coordinate of the vertex is midway between the two roots: (x = \frac{-b}{2a})
Calculator: Many calculators can find the turning point directly

For (a > 0): minimum at the vertex. For (a < 0): maximum at the vertex.

4. Worked Example: Sketching y = x² − 4x + 3

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

x-intercepts: (x^2 - 4x + 3 = 0) → ((x - 1)(x - 3) = 0) → (x = 1) or (x = 3)

Turning point: Midway between roots: (x = 2). (y = 4 - 8 + 3 = -1). Vertex at (2, −1).

Shape: (a = 1 > 0) → upright parabola ∪

5. Worked Example: Negative Quadratic

Sketch (y = -x^2 + 2x + 3)

y-intercept: (0, 3)

x-intercepts: (-x^2 + 2x + 3 = 0) → (x^2 - 2x - 3 = 0) → ((x - 3)(x + 1) = 0) → (x = 3) or (x = -1)

Turning point: (x = \frac{-b}{2a} = \frac{-2}{-2} = 1). (y = -1 + 2 + 3 = 4). Maximum at (1, 4).

Shape: (a = -1 < 0) → inverted parabola ∩

6. Examiner Tips

Your calculator may find roots and the turning point, but show your method for marks
Rearrange equations into (y = ax^2 + bx + c) when needed
Consider how many marks the question is worth to judge how much working to show