Which Function Has A Vertex At 2 9

Which Function Has a Vertex at (2, 9)? Exploring Quadratic Functions and Their Vertices

Finding a function with a specific vertex, like (2, 9), involves understanding the characteristics of different function types, particularly quadratic functions. While other functions might have points at (2, 9), the most common and straightforward scenario involves quadratic functions due to their parabolic nature and easily identifiable vertices. This article will delve into the intricacies of finding such a function, focusing on quadratic functions and exploring various approaches to solving this problem.

Understanding Quadratic Functions and Their Vertices

A quadratic function is a polynomial function of the second degree, generally represented in the form:

f(x) = ax² + bx + c

where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero (a ≠ 0). The graph of a quadratic function is a parabola, a symmetrical U-shaped curve. The vertex of the parabola is the highest or lowest point on the curve, representing either the maximum or minimum value of the function.

The x-coordinate of the vertex can be found using the formula:

x = -b / 2a

Once you have the x-coordinate, you can substitute it back into the quadratic function to find the corresponding y-coordinate, which gives you the vertex (x, y).

Finding a Quadratic Function with Vertex (2, 9)

Knowing that the vertex is (2, 9), we can use the vertex form of a quadratic function to find a suitable equation. The vertex form is expressed as:

f(x) = a(x - h)² + k

where (h, k) represents the coordinates of the vertex. In our case, (h, k) = (2, 9). Substituting these values, we get:

f(x) = a(x - 2)² + 9

Notice that 'a' remains a variable. This is because infinitely many quadratic functions can have the same vertex. The value of 'a' determines the parabola's width and whether it opens upwards (a > 0) or downwards (a < 0).

Examples with Different Values of 'a'

Let's explore a few examples with different values of 'a':

• f(x) = (x - 2)² + 9: Here, a = 1. This parabola opens upwards, has a minimum value of 9 at x = 2, and is relatively "narrow."

• f(x) = 2(x - 2)² + 9: Here, a = 2. This parabola also opens upwards, has a minimum value of 9 at x = 2, but is "narrower" than the previous example. The larger the absolute value of 'a', the narrower the parabola becomes.

• f(x) = 0.5(x - 2)² + 9: Here, a = 0.5. This parabola opens upwards, has a minimum value of 9 at x = 2, and is "wider" than the first example. The smaller the absolute value of 'a' (but still greater than 0), the wider the parabola becomes.

• f(x) = -(x - 2)² + 9: Here, a = -1. This parabola opens downwards, has a maximum value of 9 at x = 2.

• f(x) = -3(x - 2)² + 9: Here, a = -3. This parabola opens downwards, has a maximum value of 9 at x = 2, and is narrower than the previous example.

These examples demonstrate that the value of 'a' significantly influences the shape of the parabola without changing its vertex. You can choose any non-zero value for 'a' to obtain a different quadratic function with the vertex at (2, 9).

Beyond the Vertex Form: Standard Form and Completing the Square

While the vertex form offers a direct approach, you can also start with the standard form (ax² + bx + c) and use the method of completing the square to find a quadratic function with the specified vertex.

Let's say we want to find a quadratic function in standard form with a vertex at (2, 9). We know the x-coordinate of the vertex is given by -b/2a = 2. This gives us a relationship between 'a' and 'b'. However, we need more information to solve for 'a', 'b', and 'c' uniquely. We need an additional point on the parabola or some constraint to determine a unique solution.

Completing the square involves manipulating the standard form to reach the vertex form. It involves taking the standard form:

ax² + bx + c

and transforming it into:

a(x - h)² + k

where (h,k) is the vertex. This process involves several algebraic steps, including factoring out the 'a' from the x² and x terms, adding and subtracting a specific constant to create a perfect square trinomial, and then simplifying.

Illustrative Example: Completing the Square

Let's assume we are given the standard form of a quadratic equation and a point on the curve besides the vertex. Let's say we have the standard form ax² + bx + c and the vertex at (2, 9) and the additional point (0, 5).

1. Use the vertex x-coordinate: We know -b/2a = 2, which means b = -4a.

2. Substitute into the standard form: The equation is now ax² - 4ax + c.

3. Use the point (0, 5): Substitute x = 0 and f(x) = 5 into the equation: 5 = c.

4. Use the vertex (2, 9): Substitute x = 2 and f(x) = 9 into the equation: 9 = 4a - 8a + 5. This simplifies to -4a = 4, hence a = -1.

5. Find b: Since b = -4a, b = 4.

6. The complete equation: The quadratic function is -x² + 4x + 5.

You can verify that this function has a vertex at (2, 9) by either completing the square or using the formula for the x-coordinate of the vertex (-b/2a).

Other Function Types: Are There Other Possibilities?

While quadratic functions are the most straightforward way to achieve a vertex at (2, 9), other functions could potentially pass through this point. However, the concept of a "vertex" itself is most commonly associated with parabolic curves (quadratic functions). Other functions might have local maxima or minima at (2, 9), but these would not technically be considered "vertices" in the same way.

Conclusion: Infinite Solutions, Multiple Approaches

There is not a single function with a vertex at (2, 9). Instead, there are infinitely many quadratic functions that satisfy this condition. The difference lies in the value of 'a', which governs the parabola's width and whether it opens upwards or downwards. The vertex form provides the most direct method for finding such functions. The standard form necessitates additional information and the use of completing the square or similar techniques to determine a specific quadratic equation. While other functions could pass through (2, 9), the context of "vertex" strongly suggests the focus should remain on quadratic functions for a clear and precise answer. Understanding the properties of quadratic functions, particularly their vertex form, is crucial for solving problems of this nature efficiently and effectively.