Math unit 3 on 12 March 2025 at 12.12.56 PM

Understanding Quadratic Functions

• Quadratic functions can be expressed in Vertex Form:

  • Formula: A(x - h)² + k

  • Vertex: (h, k)

• Important Note: Time cannot be negative; therefore only positive values are considered.

Finding the Vertex

• To find the vertex from standard form, use the formula:

  • x = -b / (2a)

• Example:

  • Given: f(x) = 3x² + 6x + 5

  • Here, a = 3, b = 6

  • Calculate x-coordinate of vertex:

    • x = -6 / (2 * 3) = -1

• To find the y-coordinate, substitute x back into the function:

  • f(-1) = 3(-1)² + 6(-1) + 5

  • f(-1) = 3 - 6 + 5 = 2

  • Vertex is (-1, 2)

Calculator Use to Find the Vertex

• Input the function into the calculator:

  • Use the 'y = ' option for f(x)

• To find the vertex:

  • Access the graph menu, select 'Trace' and designate minimum, then navigate around the vertex to capture the values.

Analyzing the Quadratic Function's Graph

• The sign of the leading coefficient determines the orientation of the parabola:

  • Positive leading coefficient: Opens upwards and has a minimum point.

  • Negative leading coefficient: Opens downwards and has a maximum point.

• The function may have complex roots if the graph does not intersect the x-axis.

Finding X and Y Intercepts

• Finding X-Intercepts: Set f(x) = 0

  • Solve using the quadratic formula:

    • x = -b ± √(b² - 4ac) / 2a

  • Complex solutions arise if the discriminant is negative.

• Finding Y-Intercept: Set x = 0 in the function:

  • For f(x) = 3x² + 6x + 5:

    • f(0) = 5 (Y-intercept is (0, 5))

Vertex Form: Transformations and Finding Coefficient

• Example transformation:

  • New vertex at (-2, -3)

• General form: f(x) = a(x + 2)² - 3, where a is the stretch factor.

  • Solve for 'a' using another given point (e.g., (0, -1)).

  • Substitute values into the vertex equation:

    • -1 = a(0 + 2)² - 3

    • Solve for 'a' to find the exact function.

Real-World Application: Projectile Motion

• Given equation for the height of an object: h(t) = -16t² + 80t + 40

• To find maximum height:

  • Find time at maximum height (x-coordinate of the vertex):

    • t = -b / (2a) = -80 / (2 * -16) = 2.5 seconds

  • Substitute back to find maximum height:

    • h(2.5) = -16(2.5)² + 80(2.5) + 40

    • Result: 130 feet at maximum height.

Summary

• Key takeaway from quadratics; the vertex, intercepts, and maximum or minimum points provide vital information on the nature of the function's graph. Understand how to convert between forms and analyze the function both algebraically and graphically.