How To Write A Quadratic Function

Let's explore the ins and outs of writing quadratic functions, from understanding their basic structure to mastering various forms and applications.

Understanding the Basics of Quadratic Functions

A quadratic function is a polynomial function of degree two. This means the highest power of the variable in the function is two. The general form of a quadratic function is:

f(x) = ax² + bx + c

Where:

• f(x) represents the value of the function at a given value of x. It's often written as y.
• x is the independent variable.
• a, b, and c are constants, with a not equal to zero. If a were zero, the function would become linear, not quadratic.

Key Components and Their Impact:

• The coefficient 'a': This determines the direction and "width" of the parabola.

  • If a > 0, the parabola opens upwards, and the vertex is the minimum point.
  • If a < 0, the parabola opens downwards, and the vertex is the maximum point.
  • The absolute value of a affects the parabola's width. A larger absolute value makes the parabola narrower, while a smaller absolute value makes it wider.

• The coefficient 'b': This affects the position of the parabola's axis of symmetry. It influences the horizontal position of the parabola.

• The constant 'c': This represents the y-intercept of the parabola. It's the point where the parabola intersects the y-axis.

The Graph of a Quadratic Function: The Parabola

The graph of a quadratic function is a U-shaped curve called a parabola. Key features of a parabola include:

• Vertex: The highest or lowest point on the parabola. This point represents the maximum or minimum value of the function.
• Axis of Symmetry: A vertical line that passes through the vertex, dividing the parabola into two symmetrical halves. The equation of the axis of symmetry is x = -b / 2a.
• X-intercepts (Roots or Zeros): The points where the parabola intersects the x-axis. These points represent the solutions to the quadratic equation ax² + bx + c = 0.
• Y-intercept: The point where the parabola intersects the y-axis. This is found by setting x = 0 in the quadratic function, resulting in the point (0, c).

Different Forms of Quadratic Functions

Besides the general form, quadratic functions can be expressed in other forms, each providing unique insights and advantages for specific applications.

1. Standard Form (General Form):

• f(x) = ax² + bx + c

  • As discussed above, this form is straightforward and useful for identifying the coefficients a, b, and c. It directly reveals the y-intercept (c). However, finding the vertex and x-intercepts is less direct.

2. Vertex Form:

• f(x) = a(x - h)² + k

  • Advantages: This form directly reveals the vertex of the parabola, which is the point (h, k). The value of 'a' still determines the direction and width of the parabola.
  • Conversion from General Form: To convert from general form to vertex form, you need to complete the square.
    • Step 1: Factor out 'a' from the x² and x terms: f(x) = a(x² + (b/a)x) + c
    • Step 2: Complete the square inside the parentheses: Take half of the coefficient of the x term (b/2a), square it ((b/2a)²), and add and subtract it inside the parentheses: f(x) = a(x² + (b/a)x + (b/2a)² - (b/2a)²) + c
    • Step 3: Rewrite the expression inside the parentheses as a squared term: f(x) = a((x + b/2a)²) - a(b/2a)² + c
    • Step 4: Simplify: f(x) = a(x + b/2a)² + (c - ab²/4a²) Therefore, h = -b/2a and k = c - b²/4a
  • Example: Convert f(x) = 2x² + 8x + 5 to vertex form.
    • f(x) = 2(x² + 4x) + 5
    • f(x) = 2(x² + 4x + 4 - 4) + 5
    • f(x) = 2((x + 2)²) - 8 + 5
    • f(x) = 2(x + 2)² - 3 The vertex is (-2, -3).

3. Intercept Form (Factored Form):

• f(x) = a(x - r₁)(x - r₂)

  • Advantages: This form directly reveals the x-intercepts (roots or zeros) of the parabola, which are r₁ and r₂. The value of 'a' still determines the direction and width of the parabola.
  • Finding the Vertex: The x-coordinate of the vertex is the average of the x-intercepts: (r₁ + r₂) / 2. Substitute this value back into the function to find the y-coordinate of the vertex.
  • Conversion from General Form: To convert from general form to intercept form, you need to factor the quadratic expression.
    • Step 1: Set the quadratic expression equal to zero: ax² + bx + c = 0
    • Step 2: Factor the quadratic expression into two linear factors: (x - r₁)(x - r₂) = 0
    • Step 3: Write the function in intercept form: f(x) = a(x - r₁)(x - r₂)
  • Example: Convert f(x) = x² - 5x + 6 to intercept form.
    • x² - 5x + 6 = 0
    • (x - 2)(x - 3) = 0
    • f(x) = (x - 2)(x - 3) The x-intercepts are 2 and 3. The vertex x-coordinate is (2+3)/2 = 2.5. The vertex y-coordinate is (2.5-2)(2.5-3) = 0.5 * -0.5 = -0.25. So the vertex is (2.5, -0.25).

Writing Quadratic Functions from Given Information

Often, you'll be asked to write a quadratic function based on given information. The approach depends on the type of information provided.

1. Given the Vertex and a Point:

• Use Vertex Form: f(x) = a(x - h)² + k

  • Step 1: Substitute the coordinates of the vertex (h, k) into the vertex form.
  • Step 2: Substitute the coordinates of the given point (x, y) into the equation.
  • Step 3: Solve for 'a'.
  • Step 4: Write the complete quadratic function in vertex form.

• Example: Write a quadratic function with a vertex at (1, 2) and passing through the point (3, 6).

  • f(x) = a(x - 1)² + 2
  • 6 = a(3 - 1)² + 2
  • 6 = 4a + 2
  • 4a = 4
  • a = 1
  • f(x) = (x - 1)² + 2 (This can be expanded to f(x) = x² - 2x + 3 in general form)

2. Given the X-Intercepts and a Point:

• Use Intercept Form: f(x) = a(x - r₁)(x - r₂)

  • Step 1: Substitute the x-intercepts r₁ and r₂ into the intercept form.
  • Step 2: Substitute the coordinates of the given point (x, y) into the equation.
  • Step 3: Solve for 'a'.
  • Step 4: Write the complete quadratic function in intercept form.

• Example: Write a quadratic function with x-intercepts at -1 and 3 and passing through the point (1, 4).

  • f(x) = a(x + 1)(x - 3)
  • 4 = a(1 + 1)(1 - 3)
  • 4 = a(2)(-2)
  • 4 = -4a
  • a = -1
  • f(x) = -(x + 1)(x - 3) (This can be expanded to f(x) = -x² + 2x + 3 in general form)

3. Given Three Points:

• Use General Form: f(x) = ax² + bx + c

  • Step 1: Substitute the coordinates of each point (x, y) into the general form to create a system of three equations with three unknowns (a, b, and c).
  • Step 2: Solve the system of equations for a, b, and c. This can be done using substitution, elimination, or matrices.
  • Step 3: Write the complete quadratic function in general form.

• Example: Write a quadratic function passing through the points (0, -3), (1, 0), and (2, 5).

  • Using (0, -3): -3 = a(0)² + b(0) + c => c = -3
  • Using (1, 0): 0 = a(1)² + b(1) + c => a + b + c = 0
  • Using (2, 5): 5 = a(2)² + b(2) + c => 4a + 2b + c = 5
  • Substitute c = -3 into the other two equations:
    • a + b - 3 = 0 => a + b = 3
    • 4a + 2b - 3 = 5 => 4a + 2b = 8 => 2a + b = 4
  • Solve the system:
    • 2a + b = 4
    • a + b = 3
    • Subtract the second equation from the first: a = 1
    • Substitute a = 1 into a + b = 3: 1 + b = 3 => b = 2
  • Therefore, a = 1, b = 2, and c = -3.
  • f(x) = x² + 2x - 3

4. Given the Axis of Symmetry and Two Points:

• Step 1: Determine the x-coordinate of the vertex. The axis of symmetry is a vertical line that passes through the vertex, so its equation is x = h, where h is the x-coordinate of the vertex.

• Step 2: If the y-coordinate of the vertex is given, then you have the complete vertex (h, k). Use the vertex form f(x) = a(x - h)² + k and one of the given points to solve for 'a', as described in method 1 (Given the Vertex and a Point).

• Step 3: If the y-coordinate of the vertex is not given, use the general form f(x) = ax² + bx + c. You know that h = -b / 2a (the x-coordinate of the vertex). So, b = -2ah. Substitute this into the general form: f(x) = ax² - 2ahx + c.

• Step 4: Use the two given points to create two equations. Substitute the x and y values of each point into the equation from Step 3. Now you have two equations with two unknowns (a and c).

• Step 5: Solve the system of equations for a and c.

• Step 6: Substitute the values of a, b (which is -2ah), and c into the general form f(x) = ax² + bx + c to get the quadratic function.

• Example: Write a quadratic function with axis of symmetry x = 1 and passing through the points (0, 1) and (2, 5).

  • The x-coordinate of the vertex is 1. So h = 1. We don't know the y-coordinate, so we'll use the general form.
  • b = -2ah (from axis of symmetry)
  • f(x) = ax² + bx + c becomes f(x) = ax² - 2ax + c
  • Using (0, 1): 1 = a(0)² - 2a(0) + c => c = 1
  • Using (2, 5): 5 = a(2)² - 2a(2) + c => 5 = 4a - 4a + c => 5 = c
  • We have a contradiction! c cannot be both 1 and 5. This means there's no quadratic function that satisfies these conditions. We need to revisit the problem or the given information as there might be an error. Let's assume the point (2,1) instead of (2,5).

  Let's redo with the point (2,1):

  • The x-coordinate of the vertex is 1. So h = 1. We don't know the y-coordinate, so we'll use the general form.
  • b = -2ah (from axis of symmetry)
  • f(x) = ax² + bx + c becomes f(x) = ax² - 2ax + c
  • Using (0, 1): 1 = a(0)² - 2a(0) + c => c = 1
  • Using (2, 1): 1 = a(2)² - 2a(2) + c => 1 = 4a - 4a + c => 1 = c *So c = 1. Substitute into the equation: f(x) = ax² - 2ax + 1 *Now, we need another point not on the axis of symmetry to determine a uniquely. Since the function must be symmetric around x=1, let's try using the fact that (0,1) and (2,1) are symmetric around x=1.

  *Because points (0,1) and (2,1) are symmetric, knowing this is not sufficient to define the equation. Let's add the constraint (3,4) to the information given and solve as follows: f(x)= ax^2 -2ax +1. *Using (3,4): 4 = a(3)^2 - 2a(3) +1 => 3 = 9a - 6a => 3 = 3a => a=1. *Thus f(x) = x^2 - 2x +1.

Important Considerations:

• Uniqueness: Given three points, there is only one quadratic function that passes through them (unless the points are collinear, in which case it would be a linear function). However, given only the x-intercepts or the vertex, there are infinitely many quadratic functions, differing only by the value of 'a'. You need an additional point to uniquely determine the function.
• Real vs. Complex Roots: When given information leads to a quadratic equation with no real solutions, it means the parabola does not intersect the x-axis. The roots are complex numbers. This is perfectly valid, and you can still write the quadratic function, but the intercept form won't be applicable with real numbers.

Applications of Quadratic Functions

Quadratic functions have numerous applications in various fields:

• Physics: Projectile motion (the path of a thrown object) is described by a quadratic function. The height of the object as a function of time follows a parabolic trajectory.
• Engineering: Designing bridges, arches, and other structures often involves quadratic functions to ensure stability and optimal shape.
• Economics: Modeling cost, revenue, and profit functions can involve quadratic relationships. Finding the maximum profit often involves finding the vertex of a quadratic profit function.
• Computer Graphics: Parabolas are used in computer graphics to create curves and shapes.
• Optimization Problems: Many optimization problems can be modeled using quadratic functions. For example, finding the dimensions of a rectangular garden that maximize the area given a fixed perimeter can be solved using a quadratic function.

Examples of Application Problems:

• Projectile Motion: A ball is thrown upwards with an initial velocity of 20 m/s from a height of 2 meters. The height h(t) of the ball after t seconds is given by h(t) = -5t² + 20t + 2. What is the maximum height the ball reaches? When does the ball hit the ground?

  • The maximum height is the y-coordinate of the vertex. The x-coordinate (time) of the vertex is -b/2a = -20 / (2 * -5) = 2 seconds. The maximum height is h(2) = -5(2)² + 20(2) + 2 = -20 + 40 + 2 = 22 meters.
  • The ball hits the ground when h(t) = 0. -5t² + 20t + 2 = 0. Using the quadratic formula, t = (-b ± √(b² - 4ac)) / 2a = (-20 ± √(20² - 4 * -5 * 2)) / (2 * -5) = (-20 ± √(440)) / -10 = 2 ± √110 / 5. Since time cannot be negative, t = 2 + √110 / 5 ≈ 4.099 seconds.

• Maximizing Area: A farmer wants to fence off a rectangular garden next to a barn. He has 100 feet of fencing. What dimensions will maximize the area of the garden? (Assume the barn forms one side of the rectangle, so only three sides need fencing).

  • Let l be the length of the garden and w be the width. The perimeter is l + 2w = 100, so l = 100 - 2w. The area is A = l * w = (100 - 2w)w = 100w - 2w².
  • To maximize the area, we need to find the vertex of the quadratic function A(w) = -2w² + 100w. The w-coordinate of the vertex is -b/2a = -100 / (2 * -2) = 25 feet.
  • So the width is 25 feet, and the length is l = 100 - 2(25) = 50 feet. The maximum area is 25 * 50 = 1250 square feet.

Tips and Tricks for Writing Quadratic Functions

• Choose the Right Form: Select the form that best suits the given information. Vertex form is ideal when you know the vertex, intercept form when you know the x-intercepts, and general form when you have three points.
• Completing the Square: Practice completing the square to convert between general and vertex forms. This is a fundamental skill.
• Factoring: Mastering factoring techniques is crucial for converting between general and intercept forms.
• The Quadratic Formula: Remember the quadratic formula x = (-b ± √(b² - 4ac)) / 2a for finding the x-intercepts (roots) when factoring is difficult or impossible. The discriminant (b² - 4ac) tells you the nature of the roots: positive (two real roots), zero (one real root), or negative (two complex roots).
• Check Your Work: Always check your work by substituting the given information back into the quadratic function you derived. This will help you catch any errors.
• Graphing: Graphing the quadratic function (either by hand or using a graphing calculator) can provide a visual confirmation of your solution and help you understand the relationship between the function and its graph.
• Units: Pay attention to the units in application problems. Ensure that your answers have the correct units (e.g., meters, seconds, square feet).

Conclusion

Writing quadratic functions involves understanding their basic structure, mastering different forms, and applying appropriate techniques based on the given information. With practice and a solid grasp of these concepts, you can confidently tackle a wide range of problems involving quadratic functions. From projectile motion to optimization problems, the applications are vast and varied, making this a valuable skill in many fields.