Which Is F 6 For The Quadratic Function Graphed

Decoding the Vertex and Finding f(6) for a Quadratic Function

Finding the value of a quadratic function at a specific point, like f(6), requires understanding the function's characteristics. This article will guide you through identifying the quadratic function from its graph, determining its equation, and finally, calculating f(6). We'll explore various methods, from visually estimating the vertex to using algebraic techniques. This comprehensive guide will equip you with the knowledge to tackle similar problems confidently.

Understanding Quadratic Functions and Their Graphs

A quadratic function is a polynomial function of degree two, generally expressed in the form f(x) = ax² + bx + c, where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. The graph of a quadratic function is a parabola, a symmetrical U-shaped curve. The parabola opens upwards (U-shape) if 'a' is positive and downwards (inverted U-shape) if 'a' is negative.

The vertex of the parabola is the highest or lowest point on the graph, depending on whether the parabola opens downwards or upwards, respectively. The x-coordinate of the vertex is given by -b/2a. The y-coordinate is found by substituting this x-value back into the quadratic function. The vertex represents the minimum or maximum value of the function. The axis of symmetry is a vertical line that passes through the vertex, dividing the parabola into two mirror-image halves.

Identifying the vertex from a graph is crucial for determining the quadratic equation. The vertex, along with another point on the parabola, provides sufficient information to find the function's equation.

Methods for Finding the Quadratic Function from its Graph

Several methods exist to determine the quadratic function if only its graph is given:

1. Using the Vertex and Another Point:

This is arguably the most straightforward method. If the coordinates of the vertex (h, k) and another point (x₁, y₁) on the parabola are known, the quadratic function can be expressed in vertex form:

f(x) = a(x - h)² + k

Substitute the coordinates of the vertex (h, k) and the other point (x₁, y₁) into this equation. Solve for 'a', and you have the complete quadratic function.

Example: Let's say the vertex is (2, -1) and another point on the graph is (3, 2).

Substituting into the vertex form:

2 = a(3 - 2)² + (-1) 2 = a(1)² - 1 3 = a Therefore, the quadratic function is f(x) = 3(x - 2)² - 1

2. Using Three Points:

If you have the coordinates of three distinct points (x₁, y₁), (x₂, y₂), and (x₃, y₃) on the parabola, you can set up a system of three simultaneous equations:

y₁ = ax₁² + bx₁ + c y₂ = ax₂² + bx₂ + c y₃ = ax₃² + bx₃ + c

Solving this system of equations (using substitution, elimination, or matrices) will yield the values of 'a', 'b', and 'c', thus defining the quadratic function. This method is more algebraically intensive than the vertex method.

3. Using Intercept Form (if applicable):

If the x-intercepts (where the parabola crosses the x-axis) are visible on the graph, you can use the intercept form of a quadratic function:

f(x) = a(x - p)(x - q)

where 'p' and 'q' are the x-intercepts. Again, you'll need another point on the graph to determine the value of 'a'.

Calculating f(6) after Determining the Quadratic Function

Once the quadratic function f(x) is determined using any of the above methods, calculating f(6) is simply a matter of substituting x = 6 into the equation and evaluating the result. This will give you the y-coordinate of the point on the parabola where x = 6.

Example: Using the quadratic function from the previous example, f(x) = 3(x - 2)² - 1, let's find f(6):

f(6) = 3(6 - 2)² - 1 f(6) = 3(4)² - 1 f(6) = 3(16) - 1 f(6) = 48 - 1 f(6) = 47

Therefore, the y-coordinate of the point on the parabola when x = 6 is 47.

Illustrative Example with a Detailed Walkthrough

Let's work through a complete example. Suppose the graph of a quadratic function shows a vertex at (-1, 4) and passes through the point (1, 0).

1. Determine the quadratic function:

We'll use the vertex form: f(x) = a(x - h)² + k.

The vertex is (h, k) = (-1, 4). The point (x₁, y₁) = (1, 0).

Substituting:

0 = a(1 - (-1))² + 4 0 = a(2)² + 4 0 = 4a + 4 -4 = 4a a = -1

Therefore, the quadratic function is f(x) = -(x + 1)² + 4.

2. Calculate f(6):

Substitute x = 6 into the equation:

f(6) = -(6 + 1)² + 4 f(6) = -(7)² + 4 f(6) = -49 + 4 f(6) = -45

So, for this specific quadratic function, f(6) = -45.

Frequently Asked Questions (FAQ)

Q1: What if the graph doesn't clearly show the vertex or x-intercepts?

A1: If the vertex and intercepts aren't readily apparent, you'll need to estimate their coordinates as accurately as possible from the graph. The accuracy of your final answer will depend on the accuracy of your estimations. Using three distinct points, even if not precisely determined, will still give an approximation of the function.

Q2: Can I use technology to find the quadratic function?

A2: Yes, many graphing calculators and software programs (like GeoGebra or Desmos) can perform quadratic regression. Inputting the coordinates of several points on the graph will allow the software to fit a quadratic function to the data, providing a more accurate representation.

Q3: Are there other forms of quadratic equations besides vertex and intercept forms?

A3: Yes, the standard form (f(x) = ax² + bx + c) is another common representation. While you can use this form with three points, the vertex and intercept forms often simplify the calculations.

Q4: What if the parabola is very narrow or very wide? Does that affect the process?

A4: The width (or narrowness) of the parabola is determined by the value of 'a' in the quadratic equation. A larger absolute value of 'a' indicates a narrower parabola, while a smaller absolute value indicates a wider parabola. The process of finding the function and calculating f(6) remains the same, regardless of the parabola's width.

Conclusion

Finding f(6) for a quadratic function, given its graph, involves a two-step process: first, determine the quadratic function itself using either the vertex and another point, three points, or the intercepts (if visible); then, substitute x = 6 into the resulting equation to find f(6). While visual estimation can be used, utilizing algebraic methods ensures greater accuracy. Remember to check your work and ensure that the obtained quadratic function accurately reflects the given graph. This detailed explanation should equip you to handle similar problems efficiently and accurately. The key is understanding the characteristics of quadratic functions and choosing the most appropriate method based on the information available from the graph.