Quadratic Graphs

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Lesson notes

Key Features of Quadratic Graphs

A quadratic graph has equation y = ax² + bx + c with a ≠ 0.
It is a smooth curve called a parabola, with a vertical line of symmetry.
If a > 0, the graph is u-shaped (minimum turning point).
If a < 0, the graph is n-shaped (maximum turning point).
The y-intercept is at (0, c).
The x-intercepts (roots) are solutions to ax² + bx + c = 0; there can be 0, 1, or 2 roots.
The turning point (vertex) is the minimum or maximum point.

Positive and negative quadratics

Sketching a Quadratic Graph

Draw axes and mark the y-intercept (0, c).
Find and mark the roots by solving ax² + bx + c = 0 (factorising, completing square, or quadratic formula).
Determine the shape: u-shaped if a > 0, n-shaped if a < 0.
Sketch a smooth curve through the intercepts, showing the turning point if known.
Label all intercepts and the turning point coordinates.

Key features of a quadratic graph

Finding the Turning Point by Completing the Square

Rewrite y = ax² + bx + c as y = a(x - p)² + q.
The turning point is at (p, q) (note sign change for p).
For y = (x - 3)² + 2, the minimum is (3, 2).
For y = (x + 3)² + 2, the minimum is (-3, 2).
The value of a does not affect the turning point coordinates but affects the shape.

Finding the Turning Point by Differentiation

Differentiate y = ax² + bx + c to get dy/dx = 2ax + b.
Set dy/dx = 0 and solve for x to find the x-coordinate of the turning point.
Substitute this x into the original equation to find the y-coordinate.
This method works for any quadratic and confirms whether it is a maximum or minimum.

Finding the Equation of a Quadratic from Its Graph

If the vertex (p, q) and one other point are known, use y = a(x - p)² + q.
Substitute the other point to find a.
If the roots (x₁, 0) and (x₂, 0) and one other point are known, use y = a(x - x₁)(x - x₂).
Substitute the other point to find a.
If a = 1, only the vertex or roots are needed.

Example: Sketching y = x² - 5x + 6

y-intercept: (0, 6) (c = 6).
Factorise: y = (x - 2)(x - 3) → roots at (2, 0) and (3, 0).
a = 1 > 0, so graph is u-shaped.
Sketch a smooth u-shaped curve through (0,6), (2,0), (3,0).

y = x² − 5x + 6

Example: Sketching y = x² - 6x + 13

y-intercept: (0, 13).
Complete square: y = (x - 3)² + 4 → vertex at (3, 4) (minimum).
Since vertex is above x-axis and a > 0, there are no real roots.
Sketch u-shaped curve with vertex (3,4) and y-intercept (0,13).

y = x² − 6x + 13

Example: Sketching y = -x² - 4x - 4

y-intercept: (0, -4).
Differentiate: dy/dx = -2x - 4; set to 0 → x = -2.
Substitute: y = -(-2)² - 4(-2) - 4 = 0 → vertex at (-2, 0) (maximum).
Only one root at x = -2 (touches x-axis).
Sketch n-shaped curve with vertex (-2,0) and y-intercept (0,-4).

y = −x² − 4x − 4

Example: Finding Equation from Roots

Given roots at x = 2 and x = 3, and point (0, 24).
Use y = a(x - 2)(x - 3); substitute (0,24): 24 = a(-2)(-3) = 6a → a = 4.
Equation: y = 4(x - 2)(x - 3) or y = 4x² - 20x + 24.

y = 4(x − 2)(x − 3)

Example: Finding Equation from Vertex

Given vertex at (9, -16) and point (2, 82).
Use y = a(x - 9)² - 16; substitute (2,82): 82 = a(2-9)² - 16 = 49a - 16 → 49a = 98 → a = 2.
Equation: y = 2(x - 9)² - 16 or y = 2x² - 36x + 146.

y = 2(x − 9)² − 16

Slides

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Practice questions

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1.What is the shape of the graph of y = x² + 3x - 4?
Easy
- AU-shaped
- BN-shaped
- CStraight line
- DS-shaped
2.The y-intercept of y = x² - 5x + 6 is at (0, c). What is c?
Easy
- A6
- B-5
- C0
- D-6
3.How many x-intercepts can a quadratic graph have?
Easy
- A0, 1 or 2
- BAlways 2
- CAlways 1
- D0 or 2
4.The turning point of y = (x - 3)² + 5 has coordinates:
Easy
- A(3, 5)
- B(-3, 5)
- C(3, -5)
- D(-3, -5)
5.Complete the square: x² + 6x + 11 = (x + a)² + b. Find a and b.
Medium
- Aa = 3, b = 2
- Ba = -3, b = 2
- Ca = 3, b = 20
- Da = -3, b = 20
6.The graph of y = -x² + 4x - 3 has a maximum point. What is its x-coordinate?
Medium
- A2
- B-2
- C4
- D-4
7.A quadratic graph has roots at x = 1 and x = 5. Which equation could represent it?
Easy
- Ay = (x - 1)(x - 5)
- By = (x + 1)(x + 5)
- Cy = (x - 1)(x + 5)
- Dy = (x + 1)(x - 5)
8.The curve y = x² - 6x + 13 has a turning point at (3, 4). How many x-intercepts does it have?
Medium
- A0
- B1
- C2
- DCannot be determined

Revision notes

## Quadratic Graphs - Night Before Exam Summary ### Key Features - **General form**: \(y = ax^2 + bx + c\), \(a \neq 0\) - Shape: \(a > 0\) → **u-shaped** (minimum turning point); \(a < 0\) → **n-shaped** (maximum turning point) - The curve is a **parabola** with a vertical line of symmetry. - **y-intercept**: \((0, c)\) - **Roots** (x-intercepts): solutions to \(ax^2 + bx + c = 0\) (0, 1, or 2 real roots) ### Completing the Square - Write \(y = a(x - p)^2 + q\) - **Turning point** (vertex): \((p, q)\) (sign change for x!) - Example: \(y = (x-3)^2 + 2\) → minimum at \((3,2)\) - \(y = (x+3)^2 + 2\) → minimum at \((-3,2)\) - \(a\) does not affect vertex coordinates, only shape. ### Differentiation (alternative for turning point) - \(\frac{dy}{dx} = 2ax + b\) - Set \(\frac{dy}{dx} = 0\) → solve for \(x\) - Substitute back to find \(y\). ### Sketching Steps 1. Draw axes. 2. Mark y-intercept \((0,c)\). 3. Find and mark roots (if any). 4. Determine shape (u or n). 5. Find turning point (by completing square or differentiation). 6. Draw smooth curve through intercepts and vertex. ### Finding Equation from Graph - **If vertex \((p,q)\) and another point known**: use \(y = a(x-p)^2 + q\), substitute to find \(a\). - **If roots \(x_1, x_2\) and another point known**: use \(y = a(x-x_1)(x-x_2)\), substitute to find \(a\). ### Common Exam Tips - Always label intercepts and turning point coordinates. - For \(y = x^2 + 10x + 14\), complete square: \((x+5)^2 - 11\), vertex \((-5,-11)\). - For \(y = 6x - 0.5x^2 - 7\), rewrite as \(-0.5(x-6)^2 + 11\), vertex \((6,11)\), maximum. - Axis of symmetry: \(x = p\) from vertex form.

Quadratic graph: A graph of a quadratic function \(y = ax^2 + bx + c\) where \(a \neq 0\). It is a smooth curve called a parabola.
Parabola: The U-shaped or n-shaped curve of a quadratic graph.
Turning point (vertex): The minimum point (if \(a > 0\)) or maximum point (if \(a < 0\)) of a quadratic graph. Coordinates can be found by completing the square: \((p, q)\) from \(y = a(x-p)^2 + q\).
Roots: The x-values where the graph crosses the x-axis; solutions to \(ax^2 + bx + c = 0\).
y-intercept: The point where the graph crosses the y-axis; \((0, c)\).
Completing the square: Rewriting \(ax^2 + bx + c\) as \(a(x-p)^2 + q\) to find the turning point.
Axis of symmetry: The vertical line through the turning point, given by \(x = p\) in vertex form.

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