Quadratic Graphs: Foundations

[Video: Understanding Quadratic Graphs]

Learn how to interpret and analyze quadratic graphs (parabolas), including their key features and transformations.

Understanding Parabolas

Quadratic functions graph as parabolas - U-shaped curves with important features:

[Graph of a parabola showing vertex, axis of symmetry, roots, and y-intercept]

Figure 1: Key features of a quadratic graph

Standard Form: f(x) = ax² + bx + c

Vertex: The turning point of the parabola
Axis of symmetry: Vertical line through vertex (x = -b/2a)
Roots/zeros: x-intercepts (if they exist)
y-intercept: (0, c)
Direction: Opens up (a > 0) or down (a < 0)

Vertex Form and Transformations

Vertex Form: f(x) = a(x - h)² + k

Vertex: (h, k)
Axis of symmetry: x = h
a determines: Direction and width of parabola

Example: f(x) = 2(x - 3)² + 1

• Vertex at (3, 1)

• Opens upward (a = 2 > 0)

• Narrower than standard parabola (|a| > 1)

Key Graph Features

Finding Roots

Solve ax² + bx + c = 0 using:

Factoring (when possible)
Quadratic formula
Completing the square

Example: Find roots of f(x) = x² - 4x + 3

1. Set equal to 0: x² - 4x + 3 = 0

2. Factor: (x - 1)(x - 3) = 0

3. Roots: x = 1 and x = 3

Vertex Calculation

For standard form f(x) = ax² + bx + c:

x-coordinate = -b/(2a)

Then substitute x to find y-coordinate

Example: Find vertex of f(x) = -2x² + 8x - 5

1. x = -b/(2a) = -8/(2×-2) = 2

2. f(2) = -2(4) + 8(2) - 5 = 3

3. Vertex: (2, 3)

Key Takeaways

Identify key features: vertex, axis, roots, y-intercept
Understand how coefficients affect the graph's shape
Convert between standard and vertex forms when helpful
Use the vertex formula to find maxima/minima
SAT often tests interpretation of these features in context

Practice Question

The graph of the quadratic function f is shown below. If the vertex is at (2, -4) and the parabola passes through (0, 0), which of the following could be the equation of f?

[Graph showing parabola with vertex at (2, -4) passing through (0,0)]

A) f(x) = (x - 2)² - 4

B) f(x) = x² - 4x

C) f(x) = (x + 2)² - 4

D) f(x) = x² + 4x

Consider these points:

The vertex form is f(x) = a(x - h)² + k where (h,k) is vertex
Use the given point (0,0) to find the value of 'a'
Check which option satisfies both vertex and point conditions

Step-by-Step Solution:

1. Vertex form: f(x) = a(x - 2)² - 4 (from vertex (2,-4))

2. Plug in (0,0): 0 = a(0 - 2)² - 4 → 0 = 4a - 4 → a = 1

3. Equation: f(x) = (x - 2)² - 4

4. Expand to verify: f(x) = x² - 4x + 4 - 4 = x² - 4x (matches option B too)

5. Both A and B are algebraically equivalent forms of the same equation

However, option A is in vertex form which directly shows the given vertex.

The correct answer is A) f(x) = (x - 2)² - 4.