Which Equation Represents The Function Graphed On The Coordinate Plane

Which Equation Represents the Function Graphed on the Coordinate Plane? A Comprehensive Guide

Determining the equation that represents a function graphed on a coordinate plane is a fundamental skill in algebra and pre-calculus. This process involves analyzing the graph's key features, such as intercepts, slope, vertex, and asymptotes, to deduce the appropriate equation form. This comprehensive guide will walk you through various scenarios, equipping you with the knowledge and strategies to tackle diverse graphing problems.

Understanding Function Forms

Before diving into specific examples, let's review the common function forms and their characteristics:

1. Linear Functions (y = mx + b)

• Key Feature: Constant rate of change (slope). The graph is a straight line.
• m: Represents the slope (rise over run). A positive slope indicates an upward trend, while a negative slope indicates a downward trend. A slope of zero represents a horizontal line.
• b: Represents the y-intercept, the point where the line intersects the y-axis (x = 0).

Example: A line passing through (0, 2) and (1, 5). The slope is (5-2)/(1-0) = 3, and the y-intercept is 2. Therefore, the equation is y = 3x + 2.

2. Quadratic Functions (y = ax² + bx + c)

• Key Feature: Parabolic shape. The graph is a U-shaped curve that opens upwards (a > 0) or downwards (a < 0).
• Vertex: The highest or lowest point on the parabola. Its x-coordinate is given by -b/(2a).
• x-intercepts (roots): Points where the parabola intersects the x-axis (y = 0). These can be found using the quadratic formula or factoring.
• y-intercept: Point where the parabola intersects the y-axis (x = 0). This is simply the value of 'c'.

Example: A parabola with vertex (1, -4) and passing through (0, -3). Since the vertex is (h, k) = (1, -4), we can use the vertex form: y = a(x - h)² + k. Substituting the point (0, -3), we get -3 = a(0 - 1)² - 4, which solves to a = 1. Therefore, the equation is y = (x - 1)² - 4, which expands to y = x² - 2x - 3.

3. Polynomial Functions (y = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀)

• Key Feature: Can have multiple turning points (local maxima and minima). The degree of the polynomial (highest power of x) determines the maximum number of turning points (n-1).
• x-intercepts (roots): Points where the graph intersects the x-axis. The number of x-intercepts can vary depending on the degree and nature of the roots.
• y-intercept: Point where the graph intersects the y-axis (x = 0). This is simply the constant term, a₀.

Example: A cubic polynomial passing through (0, 0), (1, 0), (2, 0), and (3, 6). Since it has three x-intercepts, it can be factored as y = a(x - 1)(x - 2)(x - 0) = ax(x - 1)(x - 2). Substituting (3, 6), we get 6 = a(3)(2)(1), so a = 1. The equation is y = x(x - 1)(x - 2) = x³ - 3x² + 2x.

4. Exponential Functions (y = abˣ)

• Key Feature: Rapid growth or decay. The graph approaches a horizontal asymptote (a value the function approaches but never reaches).
• a: Represents the initial value (y-intercept when x = 0).
• b: Represents the base. If b > 1, the function represents exponential growth; if 0 < b < 1, it represents exponential decay.

Example: A graph passing through (0, 2) and (1, 6). Since it passes through (0, 2), a = 2. Using the point (1, 6), we get 6 = 2b¹, so b = 3. The equation is y = 2(3ˣ).

5. Logarithmic Functions (y = a logₓ b)

• Key Feature: Inverse of the exponential function. The graph has a vertical asymptote.
• a: Scaling factor.
• b: Base of the logarithm.

Example: Difficult to solve definitively from a graph alone without additional information like specific points or the base.

6. Rational Functions (y = p(x)/q(x))

• Key Feature: May have vertical asymptotes (where the denominator is zero) and horizontal asymptotes (determined by the degrees of the numerator and denominator).
• x-intercepts: Where the numerator is zero and the denominator is not.
• y-intercept: The value of y when x = 0 (if defined).

Example: A rational function with vertical asymptotes at x = 1 and x = -1, and a horizontal asymptote at y = 0. This suggests a denominator of (x-1)(x+1). If it passes through (0, 0), the numerator might simply be 'x'. Thus, the function could be y = x/((x-1)(x+1)).

Strategies for Determining the Equation

1. Identify the Type of Function: The first step is to visually identify the type of function based on its shape. Is it a straight line, a parabola, an exponential curve, etc.?

2. Find Key Points: Locate important points on the graph, such as intercepts, vertex, asymptotes, and other easily identifiable points.

3. Use Appropriate Equation Form: Based on the type of function, use the appropriate equation form (linear, quadratic, exponential, etc.).

4. Substitute Points: Substitute the coordinates of identified points into the equation form to solve for the unknown parameters (e.g., slope, y-intercept, vertex coordinates).

5. Verify the Equation: Once you've determined an equation, verify its accuracy by plugging in additional points from the graph.

Advanced Techniques and Considerations

• Transformations: Recognize shifts, reflections, and stretches/compressions applied to basic function graphs. These transformations change the equation accordingly. For example, y = (x-2)² + 1 shifts the parabola y = x² two units to the right and one unit up.

• System of Equations: If you have multiple unknown parameters, you may need to set up a system of equations using multiple points on the graph. Solve the system to find the parameter values.

• Technology: Graphing calculators and software can help verify your solutions and explore different equation forms.

• Contextual Information: Sometimes the problem will provide additional information about the function, such as its domain or range, which can help narrow down the possibilities.

Practice Problems

Let's work through a few examples:

Problem 1: A linear function passes through the points (-1, 2) and (1, 6). Find the equation.

Solution: The slope is (6 - 2) / (1 - (-1)) = 2. Using the point-slope form, y - 2 = 2(x + 1), which simplifies to y = 2x + 4.

Problem 2: A parabola passes through (0, 3), (1, 0), and (2, -1). Find the equation.

Solution: This is a bit more challenging and requires substituting these points into the general quadratic form y = ax² + bx + c. This gives you a system of three equations:

• 3 = c
• 0 = a + b + c
• -1 = 4a + 2b + c

Substitute c = 3 into the second and third equations to get:

• 0 = a + b + 3
• -1 = 4a + 2b + 3

Solving this system yields a = -2 and b = 1. Thus, the equation is y = -2x² + x + 3.

Problem 3: An exponential function passes through (0, 1) and (1, 3). Find the equation.

Solution: Since the function passes through (0, 1), the initial value a = 1. Using the point (1, 3), we have 3 = 1 * b¹, so b = 3. The equation is y = 3ˣ.

By mastering the techniques outlined in this guide, you'll be well-equipped to confidently tackle any graph-to-equation problem. Remember to carefully analyze the graph's characteristics, select the appropriate equation form, and systematically solve for the unknown parameters. With practice, this skill will become second nature.