Write An Equation Any Form For The Quadratic Graphed Below

The graph of a quadratic function unveils a treasure trove of information, allowing us to construct its equation in various forms. Each form offers a unique perspective on the function's behavior and key features. This article will guide you through the process of writing an equation for a quadratic function based on its graph, exploring different forms and providing practical examples.

Understanding Quadratic Functions

A quadratic function is a polynomial function of degree two, generally expressed in the following forms:

• Standard Form: f(x) = ax² + bx + c, where a, b, and c are constants and a ≠ 0. This form is useful for identifying the y-intercept (c) and determining the direction the parabola opens (a).
• Vertex Form: f(x) = a(x - h)² + k, where (h, k) represents the vertex of the parabola. This form is particularly convenient when the vertex is known.
• Factored Form (Intercept Form): f(x) = a(x - r₁)(x - r₂), where r₁ and r₂ are the x-intercepts (roots or zeros) of the function. This form is useful when the x-intercepts are easily identifiable.

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

• Vertex: The point where the parabola changes direction (either the minimum or maximum point).
• Axis of Symmetry: A vertical line that passes through the vertex, dividing the parabola into two symmetrical halves. Its equation is x = h, where h is the x-coordinate of the vertex.
• x-intercepts (Roots/Zeros): The points where the parabola intersects the x-axis (where f(x) = 0).
• y-intercept: The point where the parabola intersects the y-axis (where x = 0).

Extracting Information from the Graph

Before attempting to write the equation, carefully examine the graph and identify the following information:

1. Vertex: Determine the coordinates (h, k) of the vertex. This is the most crucial piece of information for using the vertex form.
2. x-intercepts: Identify the points where the parabola crosses the x-axis. These are the roots r₁ and r₂, essential for the factored form.
3. y-intercept: Locate the point where the parabola crosses the y-axis. This gives you the value of c in the standard form, or a point (0, y) that can be used to solve for a in other forms.
4. Any Other Point: If the vertex, x-intercepts, and y-intercept aren't readily available or sufficient, identify another clear point (x, y) on the graph. This point will be used to solve for the leading coefficient a.
5. Direction of Opening: Determine whether the parabola opens upwards (a > 0) or downwards (a < 0). This provides the sign of the leading coefficient a.

Writing the Equation in Vertex Form

The vertex form, f(x) = a(x - h)² + k, is particularly useful when the vertex (h, k) is known.

Steps:

1. Identify the Vertex: From the graph, determine the coordinates of the vertex (h, k).
2. Substitute h and k: Plug the values of h and k into the vertex form equation: f(x) = a(x - h)² + k. Now, only a remains unknown.
3. Find a Point on the Graph (Other than the Vertex): Choose any other point (x, y) on the graph that is easily readable. The y-intercept is often a good choice if it's clear.
4. Substitute x and y: Plug the x and y values of the chosen point into the equation f(x) = a(x - h)² + k. This leaves you with an equation with only a as the unknown.
5. Solve for a: Solve the equation for a. This will give you the leading coefficient.
6. Write the Complete Equation: Substitute the values of a, h, and k back into the vertex form equation f(x) = a(x - h)² + k.

Example:

Suppose the graph shows a parabola with a vertex at (2, -1) and passes through the point (0, 3).

1. Vertex: (h, k) = (2, -1)
2. Substitute: f(x) = a(x - 2)² - 1
3. Point: (x, y) = (0, 3)
4. Substitute: 3 = a(0 - 2)² - 1
5. Solve for a:
   • 3 = a(-2)² - 1
   • 3 = 4a - 1
   • 4 = 4a
   • a = 1
6. Complete Equation: f(x) = 1(x - 2)² - 1 or f(x) = (x - 2)² - 1

Writing the Equation in Factored Form (Intercept Form)

The factored form, f(x) = a(x - r₁)(x - r₂), is most convenient when the x-intercepts r₁ and r₂ are clearly visible on the graph.

Steps:

1. Identify the x-intercepts: From the graph, determine the x-intercepts r₁ and r₂. These are the points where the parabola crosses the x-axis.
2. Substitute r₁ and r₂: Plug the values of r₁ and r₂ into the factored form equation: f(x) = a(x - r₁)(x - r₂). Now, only a remains unknown.
3. Find a Point on the Graph (Other than the x-intercepts): Choose any other point (x, y) on the graph that is easily readable. The y-intercept is often a good choice if it's clear.
4. Substitute x and y: Plug the x and y values of the chosen point into the equation f(x) = a(x - r₁)(x - r₂). This leaves you with an equation with only a as the unknown.
5. Solve for a: Solve the equation for a. This will give you the leading coefficient.
6. Write the Complete Equation: Substitute the values of a, r₁, and r₂ back into the factored form equation f(x) = a(x - r₁)(x - r₂).

Example:

Suppose the graph shows a parabola with x-intercepts at (-1, 0) and (3, 0), and passes through the point (0, -3).

1. x-intercepts: r₁ = -1, r₂ = 3
2. Substitute: f(x) = a(x - (-1))(x - 3) or f(x) = a(x + 1)(x - 3)
3. Point: (x, y) = (0, -3)
4. Substitute: -3 = a(0 + 1)(0 - 3)
5. Solve for a:
   • -3 = a(1)(-3)
   • -3 = -3a
   • a = 1
6. Complete Equation: f(x) = 1(x + 1)(x - 3) or f(x) = (x + 1)(x - 3)

Writing the Equation in Standard Form

While the standard form, f(x) = ax² + bx + c, isn't the easiest to directly derive from a graph, you can obtain it by expanding either the vertex form or the factored form after you've found the equation in one of those forms.

Method 1: Expanding Vertex Form

1. Find the Equation in Vertex Form: Use the method described above to find the equation in vertex form, f(x) = a(x - h)² + k.
2. Expand: Expand the squared term and simplify the equation.

Example:

Using the vertex form equation from the previous example: f(x) = (x - 2)² - 1

1. Expand:
   • f(x) = (x - 2)(x - 2) - 1
   • f(x) = x² - 4x + 4 - 1
   • f(x) = x² - 4x + 3

Method 2: Expanding Factored Form

1. Find the Equation in Factored Form: Use the method described above to find the equation in factored form, f(x) = a(x - r₁)(x - r₂).
2. Expand: Multiply the binomials and simplify the equation.

Example:

Using the factored form equation from the previous example: f(x) = (x + 1)(x - 3)

1. Expand:
   • f(x) = x² - 3x + x - 3
   • f(x) = x² - 2x - 3

Method 3: Using a System of Equations

This method is useful when you have three distinct points on the graph but neither the vertex nor the x-intercepts are readily apparent.

1. Identify Three Points: Choose three clear points (x₁, y₁), (x₂, y₂), and (x₃, y₃) on the graph.
2. Substitute into Standard Form: Substitute each point into the standard form equation f(x) = ax² + bx + c to create three equations:
   • y₁ = ax₁² + bx₁ + c
   • y₂ = ax₂² + bx₂ + c
   • y₃ = ax₃² + bx₃ + c
3. Solve the System of Equations: Solve the resulting system of three equations for a, b, and c. You can use substitution, elimination, or matrix methods.
4. Write the Complete Equation: Substitute the values of a, b, and c back into the standard form equation f(x) = ax² + bx + c.

Example:

Suppose the graph passes through the points (1, 2), (2, 3), and (-1, 6).

1. Points: (1, 2), (2, 3), and (-1, 6)
2. Substitute:
   • 2 = a(1)² + b(1) + c => 2 = a + b + c
   • 3 = a(2)² + b(2) + c => 3 = 4a + 2b + c
   • 6 = a(-1)² + b(-1) + c => 6 = a - b + c
3. Solve the System: Solving this system (using substitution or elimination) yields a = 1, b = -2, and c = 3.
4. Complete Equation: f(x) = 1x² - 2x + 3 or f(x) = x² - 2x + 3

Choosing the Right Form

The best form to use depends on the information you can readily extract from the graph:

• Vertex Form: Ideal when the vertex is easily identifiable.
• Factored Form: Ideal when the x-intercepts are easily identifiable.
• Standard Form: Useful if you have three distinct points and other methods are cumbersome, or when you need to identify the y-intercept quickly (which is simply c).

It's important to remember that once you have the equation in one form, you can always convert it to another form through algebraic manipulation.

Common Mistakes to Avoid

• Incorrectly Identifying the Vertex or Intercepts: Double-check the coordinates of the vertex and intercepts. A small error here can significantly impact the equation.
• Forgetting the Leading Coefficient a: Don't assume a = 1. Always solve for a using another point on the graph. The value of 'a' determines whether the graph opens upwards or downwards and also affects the width of the parabola.
• Sign Errors: Be careful with signs, especially when substituting values into the equations.
• Algebraic Errors: Double-check your algebraic manipulations when solving for a or expanding the equations.
• Assuming Symmetry: While parabolas are symmetrical, don't assume a point exists without verifying it on the graph. Use the axis of symmetry to help you, but not as your only source of information.

Examples with Detailed Solutions

Example 1:

The graph shows a parabola with a vertex at (-1, 4) and passing through the point (0, 3).

1. Vertex: (h, k) = (-1, 4)
2. Vertex Form: f(x) = a(x - (-1))² + 4 => f(x) = a(x + 1)² + 4
3. Point: (x, y) = (0, 3)
4. Substitute: 3 = a(0 + 1)² + 4
5. Solve for a:
   • 3 = a(1)² + 4
   • 3 = a + 4
   • a = -1
6. Complete Vertex Form: f(x) = -1(x + 1)² + 4 or f(x) = -(x + 1)² + 4
7. Standard Form (Expanding):
   • f(x) = -(x² + 2x + 1) + 4
   • f(x) = -x² - 2x - 1 + 4
   • f(x) = -x² - 2x + 3

Example 2:

The graph shows a parabola with x-intercepts at (-2, 0) and (4, 0), and passing through the point (0, 8).

1. x-intercepts: r₁ = -2, r₂ = 4
2. Factored Form: f(x) = a(x - (-2))(x - 4) => f(x) = a(x + 2)(x - 4)
3. Point: (x, y) = (0, 8)
4. Substitute: 8 = a(0 + 2)(0 - 4)
5. Solve for a:
   • 8 = a(2)(-4)
   • 8 = -8a
   • a = -1
6. Complete Factored Form: f(x) = -1(x + 2)(x - 4) or f(x) = -(x + 2)(x - 4)
7. Standard Form (Expanding):
   • f(x) = -(x² - 4x + 2x - 8)
   • f(x) = -(x² - 2x - 8)
   • f(x) = -x² + 2x + 8

Example 3:

The graph passes through the points (1, -1), (2, -3), and (3, -5).

1. Points: (1, -1), (2, -3), and (3, -5)
2. Substitute:
   • -1 = a(1)² + b(1) + c => -1 = a + b + c
   • -3 = a(2)² + b(2) + c => -3 = 4a + 2b + c
   • -5 = a(3)² + b(3) + c => -5 = 9a + 3b + c
3. Solve the System: Solving this system (using substitution or elimination) yields a = 0, b = -1, and c = 0.
4. Complete Standard Form: f(x) = 0x² - 1x + 0 or f(x) = -x

(Note: This example resulted in a linear function, not a quadratic. This could happen if the three points provided are collinear.)

Conclusion

Writing an equation for a quadratic function from its graph involves carefully extracting key features and applying the appropriate form. By understanding the relationships between the graph and the different forms of quadratic equations, you can confidently determine the equation that represents the given parabola. Remember to double-check your work and avoid common mistakes to ensure accuracy. Practice with various examples will solidify your understanding and improve your skills in working with quadratic functions.