Applications of quadratic functions

Applications of Quadratic Functions

Quadratic functions are polynomial functions of degree two. They can be expressed in the standard form:

f(x) = ax² + bx + c

where a, b, and c are constants, and a ≠ 0. Quadratic functions have many applications in real life, such as in physics, engineering, and economics. In this topic, we will explore how to apply quadratic functions to solve various problems.

Identifying Quadratic Functions

A quadratic function can be identified by its characteristic U-shaped graph called a parabola. The direction of the parabola depends on the sign of the coefficient a:

• If a > 0, the parabola opens upwards.
• If a < 0, the parabola opens downwards.

For example, consider the quadratic function:

f(x) = 2x² - 4x + 1

Here, a = 2 (positive), so the parabola opens upwards.

Finding the Vertex

The vertex of a parabola is the highest or lowest point on the graph, depending on the direction it opens. The x-coordinate of the vertex can be found using the formula:

x = -b / (2a)

Using our example function:

f(x) = 2x² - 4x + 1

Here, a = 2 and b = -4. Plugging these values into the vertex formula:

x = -(-4) / (2 × 2) = 4 / 4 = 1

Now, substitute x = 1 back into the function to find the y-coordinate of the vertex:

f(1) = 2(1)² - 4(1) + 1 = 2 - 4 + 1 = -1

The vertex of the function is at the point (1, -1).

Applications in Real Life

Quadratic functions can model various real-life situations. Here are a few examples:

1. Projectile Motion

When an object is thrown into the air, its height can be modeled by a quadratic function. For instance, if a ball is thrown upwards with an initial velocity, the height h (in metres) after t seconds can be described by the equation:

h(t) = -4.9t² + vt + h₀

where v is the initial velocity and h₀ is the initial height. The term -4.9t² accounts for the effect of gravity.

Example: Ball Thrown Upwards

Suppose a ball is thrown upwards with an initial velocity of 20 m/s from a height of 1.5 m. The height function will be:

h(t) = -4.9t² + 20t + 1.5

To find out when the ball reaches its maximum height, we first find the vertex:

x = -b / (2a) = -20 / (2 × -4.9) = -20 / -9.8 ≈ 2.04 seconds

Now substitute t = 2.04 back into the height function:

h(2.04) = -4.9(2.04)² + 20(2.04) + 1.5

Calculating this gives:

h(2.04) ≈ -4.9(4.16) + 40.8 + 1.5 ≈ -20.384 + 40.8 + 1.5 ≈ 21.916 m

The maximum height reached by the ball is approximately 21.92 m.

2. Profit Maximisation

In business, quadratic functions can model profit based on the number of items sold. For example, if the profit P (in rand) from selling x items is given by:

P(x) = -5x² + 100x - 200

Here, a = -5 (indicating a downward-opening parabola), b = 100, and c = -200. To find the number of items that maximises profit, we find the vertex:

x = -b / (2a) = -100 / (2 × -5) = 100 / 10 = 10

Now, substitute x = 10 back into the profit function:

P(10) = -5(10)² + 100(10) - 200

This simplifies to:

P(10) = -500 + 1000 - 200 = 300 rand

The maximum profit occurs when 10 items are sold, yielding a profit of 300 rand.

Solving Quadratic Equations

Quadratic equations can be solved using various methods, including factoring, completing the square, and the quadratic formula. The quadratic formula is:

x = (-b ± √(b² - 4ac)) / (2a)

The expression under the square root, b² - 4ac, is called the discriminant. It determines the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is one real root (a double root).
• If it is negative, there are no real roots (the solutions are complex).

Example: Solving a Quadratic Equation

Consider the equation:

2x² - 4x - 6 = 0

Here, a = 2, b = -4, and c = -6. First, calculate the discriminant:

D = b² - 4ac = (-4)² - 4(2)(-6) = 16 + 48 = 64

Since D is positive, there will be two distinct real roots. Now, apply the quadratic formula:

x = (4 ± √64) / (2 × 2) = (4 ± 8) / 4

This gives us two solutions:

x₁ = (4 + 8) / 4 = 12 / 4 = 3

x₂ = (4 - 8) / 4 = -4 / 4 = -1

The solutions to the equation 2x² - 4x - 6 = 0 are x = 3 and x = -1.

Graphing Quadratic Functions

Graphing a quadratic function involves plotting the vertex and the y-intercept, and drawing the parabola. The y-intercept can be found by evaluating f(0):

f(0) = c

For the function f(x) = 2x² - 4x + 1, the y-intercept is:

f(0) = 1

To graph the function, plot the vertex (1, -1) and the y-intercept (0, 1). Then, calculate additional points by selecting x-values and finding their corresponding f(x) values. For example:

f(-1) = 2(-1)² - 4(-1) + 1 = 2 + 4 + 1 = 7

f(2) = 2(2)² - 4(2) + 1 = 8 - 8 + 1 = 1

Plot these points and draw a smooth curve through them to form the parabola.

Summary

Quadratic functions are useful in various real-life applications, including projectile motion and profit maximisation. Understanding how to find the vertex, solve quadratic equations, and graph these functions is essential. Remember to use the quadratic formula and check your solutions.

Check your understanding

1. What is the standard form of a quadratic function?
2. How do you find the vertex of a parabola?
3. What does the discriminant tell you about the roots of a quadratic equation?
4. Provide an example of a real-life situation that can be modeled by a quadratic function.