Finding The Quadratic Function Equation With Minimum At (7,-3) And Passing Through (9,9)
In the realm of mathematics, quadratic functions hold a significant position, characterized by their parabolic curves and unique properties. Understanding the equation of a quadratic function is crucial for various applications, from modeling projectile motion to optimizing engineering designs. This article delves into the process of determining the equation of a quadratic function given its minimum point and another point it passes through, providing a comprehensive guide for students and enthusiasts alike.
Understanding Quadratic Functions
Before we embark on the quest to find the equation, let's first establish a solid understanding of quadratic functions. 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's orientation (opening upwards or downwards) is determined by the sign of the coefficient 'a'. If 'a' is positive, the parabola opens upwards, indicating a minimum point. Conversely, if 'a' is negative, the parabola opens downwards, indicating a maximum point.
The vertex of the parabola is the point where the function reaches its minimum or maximum value. The x-coordinate of the vertex is given by -b/2a, and the y-coordinate is the function's value at that x-coordinate. The axis of symmetry is a vertical line passing through the vertex, dividing the parabola into two symmetrical halves.
Vertex Form of a Quadratic Function
While the standard form (ax² + bx + c) is useful for certain calculations, the vertex form provides a more direct representation of the vertex and the parabola's shape. The vertex form is expressed as:
f(x) = a(x - h)² + k
where (h, k) represents the coordinates of the vertex. This form allows us to readily identify the vertex and the vertical stretch or compression factor 'a'.
Determining the Equation: A Step-by-Step Approach
Now, let's tackle the problem at hand: finding the equation of the quadratic function with a minimum at (7, -3) and passing through the point (9, 9). We will employ the vertex form of the quadratic function to solve this problem.
Step 1: Utilize the Vertex Form
Since we are given the minimum point (7, -3), which corresponds to the vertex of the parabola, we can directly substitute these values into the vertex form:
f(x) = a(x - 7)² + (-3)
Simplifying, we get:
f(x) = a(x - 7)² - 3
Step 2: Incorporate the Additional Point
We are also given that the function passes through the point (9, 9). This means that when x = 9, f(x) = 9. We can substitute these values into the equation we derived in Step 1:
9 = a(9 - 7)² - 3
Step 3: Solve for 'a'
Now, we have an equation with only one unknown, 'a'. Let's solve for it:
9 = a(2)² - 3
9 = 4a - 3
12 = 4a
a = 3
Step 4: Complete the Equation
We have now determined the value of 'a' to be 3. We can substitute this value back into the equation from Step 1 to obtain the complete equation of the quadratic function:
f(x) = 3(x - 7)² - 3
Expanding the Equation (Optional)
While the vertex form is perfectly valid, we can also expand the equation to express it in the standard form (ax² + bx + c). Let's expand the equation we derived:
f(x) = 3(x² - 14x + 49) - 3
f(x) = 3x² - 42x + 147 - 3
f(x) = 3x² - 42x + 144
Therefore, the equation of the quadratic function in standard form is:
f(x) = 3x² - 42x + 144
Verification and Conclusion
To ensure our solution is correct, we can verify that the vertex is indeed at (7, -3) and that the point (9, 9) lies on the parabola. We can also graph the function to visually confirm its properties.
In conclusion, we have successfully determined the equation of the quadratic function with a minimum at (7, -3) and passing through (9, 9). The equation in vertex form is f(x) = 3(x - 7)² - 3, and the equation in standard form is f(x) = 3x² - 42x + 144. This process demonstrates the power of the vertex form in solving quadratic function problems.
Further Exploration and Applications
The concepts explored in this article form the foundation for various mathematical and real-world applications. Understanding quadratic functions is essential for:
- Modeling projectile motion: The trajectory of a projectile, such as a ball thrown in the air, can be accurately modeled using a quadratic function.
- Optimization problems: Quadratic functions can be used to find the maximum or minimum values in various scenarios, such as maximizing profit or minimizing cost.
- Engineering design: Engineers use quadratic functions to design structures, bridges, and other systems that require optimal performance.
- Curve fitting: Quadratic functions can be used to fit curves to data points, allowing us to analyze trends and make predictions.
By delving deeper into the properties and applications of quadratic functions, students and enthusiasts can unlock a powerful tool for problem-solving and analysis in various fields.
Practice Problems
To solidify your understanding, try solving the following practice problems:
- Find the equation of the quadratic function with a maximum at (-2, 5) and passing through the point (0, 1).
- Determine the equation of the quadratic function with a vertex at (3, -1) and an x-intercept at (1, 0).
- A parabola has a minimum at (4, -2) and passes through the point (6, 2). Find its equation.
By working through these problems, you will gain confidence in your ability to determine the equation of a quadratic function given various pieces of information.
In conclusion, mastering the concepts and techniques presented in this article will empower you to confidently tackle quadratic function problems and unlock their vast applications in mathematics and beyond. Remember to practice regularly and explore different problem-solving approaches to enhance your understanding and skills. The world of quadratic functions awaits your exploration!
