Finding The Standard Form Of F(x) = -9(x+5)^2 + 4

Let's dive into the world of quadratic functions and explore how to transform a given function into its standard form. In this article, we'll dissect the process of converting a quadratic function from vertex form to standard form, using a specific example to illustrate the steps involved. The question we're tackling is: Which of the following is the standard form of the function $f(x) = -9(x+5)^2 + 4$? Understanding the standard form is crucial for various applications, including finding the vertex, axis of symmetry, and roots of the quadratic equation. The standard form of a quadratic function is expressed as $f(x) = ax^2 + bx + c$, where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. This form provides a clear view of the quadratic's coefficients, which directly influence its graph and properties. Now, let's embark on the journey of converting the given function into this standard form.

The standard form of a quadratic equation is a fundamental concept in algebra, providing a structured way to represent these parabolic functions. It is defined as $f(x) = ax^2 + bx + c$, where a, b, and c are constants, and crucially, a cannot be zero. The coefficient a dictates the parabola's direction (upward if positive, downward if negative) and its width. The coefficient b influences the parabola's horizontal position, while c represents the y-intercept, the point where the parabola crosses the vertical axis. This form is incredibly versatile, allowing us to easily identify key features of the parabola. For instance, the y-intercept is immediately apparent as the constant term c. Further analysis, like using the quadratic formula or completing the square, can reveal the x-intercepts (roots) and the vertex of the parabola. Moreover, the standard form is essential for solving quadratic equations and tackling real-world problems modeled by quadratic functions, such as projectile motion or optimization scenarios. Understanding this form empowers us to analyze and manipulate quadratic equations effectively. So, mastering the standard form is a cornerstone for success in algebra and its applications.

The importance of standard form in quadratic functions cannot be overstated. It's the gateway to unlocking a wealth of information about the parabola represented by the function. The standard form, $f(x) = ax^2 + bx + c$, acts as a blueprint, clearly displaying the coefficients that govern the parabola's shape, position, and orientation. The leading coefficient, a, is particularly insightful. Its sign immediately tells us whether the parabola opens upwards (if positive) or downwards (if negative). The magnitude of a also dictates the parabola's width – a larger absolute value results in a narrower parabola, while a smaller value creates a wider one. The constant term, c, is a direct indicator of the y-intercept, the point where the parabola intersects the y-axis. This is a crucial point for graphing and understanding the function's behavior. While the standard form readily reveals the y-intercept, finding the x-intercepts (roots) and the vertex requires further manipulation. Techniques like the quadratic formula or completing the square, which are often applied to the standard form, allow us to determine these key features. Furthermore, the standard form is not just a theoretical construct; it's a practical tool for solving quadratic equations and modeling real-world phenomena. From physics to engineering to economics, quadratic functions and their standard form play a vital role in representing and analyzing various systems and problems. Therefore, a solid grasp of the standard form is essential for anyone delving into the world of quadratic functions.

Converting from Vertex Form to Standard Form

To convert the given function, $f(x) = -9(x+5)^2 + 4$, from vertex form to standard form, we need to expand and simplify the expression. The vertex form provides a quick way to identify the vertex of the parabola, but the standard form helps in understanding the coefficients and other properties. Our first step involves expanding the squared term: $(x+5)^2$. This expansion can be done using the FOIL method (First, Outer, Inner, Last) or by recognizing the perfect square trinomial pattern. Either way, we get: $(x+5)^2 = x^2 + 10x + 25$. Next, we substitute this back into our original function: $f(x) = -9(x^2 + 10x + 25) + 4$. Now, we distribute the -9 across the terms inside the parentheses: $f(x) = -9x^2 - 90x - 225 + 4$. Finally, we combine the constant terms: $f(x) = -9x^2 - 90x - 221$. This final expression is in the standard form $f(x) = ax^2 + bx + c$, where a = -9, b = -90, and c = -221. This transformation allows us to easily see the coefficients of the quadratic function, which are useful for various calculations and analyses.

The process of converting from vertex form to standard form is a crucial skill in algebra. It allows us to bridge the gap between two different representations of the same quadratic function, each offering unique insights. The vertex form, $f(x) = a(x-h)^2 + k$, immediately reveals the vertex of the parabola at the point (h, k). However, it doesn't directly showcase the coefficients in the standard form, $f(x) = ax^2 + bx + c$, which are essential for other calculations and analyses. The conversion involves a systematic expansion and simplification. The first key step is to expand the squared term, $(x-h)^2$. This typically involves using the FOIL method (First, Outer, Inner, Last) or recognizing the perfect square trinomial pattern. Once expanded, the expression is substituted back into the original function. The next step is distributing the coefficient a across all terms resulting from the expansion. This ensures that each term is properly scaled according to the leading coefficient. Finally, the constant terms are combined to arrive at the standard form. This process not only transforms the function's appearance but also makes the coefficients readily accessible. These coefficients, a, b, and c, play significant roles in determining the parabola's shape, position, and intercepts. Therefore, mastering this conversion is essential for a comprehensive understanding of quadratic functions.

The detailed steps involved in converting from vertex form to standard form may seem intricate at first, but with practice, they become second nature. Let's break down each step to ensure clarity. Starting with the vertex form, $f(x) = a(x-h)^2 + k$, the initial focus is on the squared term, $(x-h)^2$. This requires expansion, and the most common methods are the FOIL method or recognizing the perfect square trinomial pattern. The FOIL method systematically multiplies the terms: First, Outer, Inner, Last. For instance, $(x-h)^2$ expands to $(x-h)(x-h) = x^2 - hx - hx + h^2 = x^2 - 2hx + h^2$. Alternatively, recognizing the perfect square trinomial pattern, $(a - b)^2 = a^2 - 2ab + b^2$, directly leads to the same result. Once the squared term is expanded, the resulting expression is substituted back into the original function. The next crucial step is distributing the coefficient a across all terms inside the parentheses. This ensures that each term is scaled appropriately by the leading coefficient. For example, if we have $f(x) = a(x^2 - 2hx + h^2) + k$, distributing a gives us $f(x) = ax^2 - 2ahx + ah^2 + k$. Finally, the constant terms, $ah^2$ and k, are combined to simplify the expression. This results in the standard form, $f(x) = ax^2 + bx + c$, where b is -2ah and c is $ah^2 + k$. Each step, from expanding the squared term to combining constants, is crucial for accurately converting the function. Mastery of these steps empowers you to manipulate quadratic functions effectively and understand their properties.

Solution and Explanation

Applying the conversion process to our given function, $f(x) = -9(x+5)^2 + 4$, we've already expanded and simplified the expression. Let's revisit the steps. First, we expanded $(x+5)^2$ to get $x^2 + 10x + 25$. Then, we substituted this back into the function: $f(x) = -9(x^2 + 10x + 25) + 4$. Next, we distributed the -9: $f(x) = -9x^2 - 90x - 225 + 4$. Finally, we combined the constant terms: $f(x) = -9x^2 - 90x - 221$. Comparing this result with the given options, we see that it matches option C: $f(x) = -9x^2 - 90x - 221$. Therefore, the standard form of the given function is $-9x^2 - 90x - 221$. This process highlights the importance of careful expansion and simplification when converting between different forms of quadratic functions. Understanding each step ensures accuracy and allows for a deeper comprehension of the function's properties.

The solution to our problem clearly demonstrates the power of the conversion process. By systematically expanding and simplifying the given function, $f(x) = -9(x+5)^2 + 4$, we successfully transformed it into its standard form. The steps, which involved expanding the squared term, distributing the coefficient, and combining constants, are the cornerstones of this conversion. Each step played a crucial role in revealing the coefficients of the standard form. The expansion of $(x+5)^2$ was the initial step, transforming the squared term into a trinomial. Distributing the -9 across the terms of the trinomial ensured that each term was correctly scaled. Finally, combining the constant terms brought us to the standard form: $f(x) = -9x^2 - 90x - 221$. This final form allows us to easily identify the coefficients: a = -9, b = -90, and c = -221. The negative value of a indicates that the parabola opens downwards, while the other coefficients influence its position and shape. This solution not only provides the answer but also reinforces the importance of meticulous execution in algebraic manipulations. A single error in any step could lead to an incorrect result. Therefore, a clear understanding of the process and careful attention to detail are paramount for success in such conversions.

The explanation behind the solution is as important as the solution itself. It provides the reasoning and justification for each step, fostering a deeper understanding of the underlying concepts. In our case, the explanation revolves around the principles of algebraic manipulation and the properties of quadratic functions. Starting with the vertex form, $f(x) = -9(x+5)^2 + 4$, we aimed to transform it into the standard form, $f(x) = ax^2 + bx + c$. The explanation highlights why each step is necessary. Expanding the squared term, $(x+5)^2$, is crucial because it removes the parentheses and exposes the individual terms. Distributing the -9 is essential for scaling each term according to the leading coefficient. Combining the constants simplifies the expression and brings it closer to the standard form. The explanation also emphasizes the significance of the resulting coefficients. The coefficient a, which is -9 in our case, determines the parabola's direction and width. The coefficient b, -90, influences the parabola's horizontal position, and the constant term c, -221, represents the y-intercept. By understanding the explanation, we gain more than just the answer; we gain a comprehensive understanding of how and why the solution works. This deeper understanding is invaluable for tackling future problems and applying these concepts in various contexts. Therefore, always striving to understand the explanation behind a solution is key to mastering mathematical concepts.

Conclusion

In conclusion, the standard form of the quadratic function $f(x) = -9(x+5)^2 + 4$ is $f(x) = -9x^2 - 90x - 221$, which corresponds to option C. This conversion process underscores the importance of algebraic manipulation skills and the ability to recognize and apply standard forms. Understanding the standard form of a quadratic function is crucial for analyzing its properties, graphing, and solving related problems. By mastering these techniques, you'll be well-equipped to tackle a wide range of quadratic function challenges.

The final answer, which is option C, represents the culmination of our conversion process. It's not just a solution; it's a testament to the power of algebraic manipulation and the importance of understanding different forms of quadratic functions. The journey from vertex form to standard form involved several key steps, each contributing to the final result. Expanding the squared term, distributing the coefficient, and combining constants were the pivotal actions that led us to the standard form: $f(x) = -9x^2 - 90x - 221$. This final form provides a clear view of the function's coefficients, which are essential for various analyses. The leading coefficient, -9, indicates a downward-opening parabola. The other coefficients, -90 and -221, further define the parabola's shape and position. The correct identification of option C as the standard form demonstrates a firm grasp of the conversion process and the properties of quadratic functions. It also highlights the importance of accuracy and attention to detail in algebraic manipulations. A single error in any step could lead to an incorrect answer. Therefore, the final answer is not just a solution; it's a validation of the skills and knowledge applied throughout the process.

Mastering the conversion between different forms of quadratic functions, such as the vertex form and standard form, is a cornerstone of algebraic proficiency. It's not just about following a set of steps; it's about understanding the underlying principles and the implications of each transformation. The standard form, $f(x) = ax^2 + bx + c$, is a versatile representation that reveals key information about the parabola. The coefficient a dictates the parabola's direction and width, c represents the y-intercept, and further analysis can uncover the x-intercepts and vertex. The vertex form, $f(x) = a(x-h)^2 + k$, on the other hand, directly displays the vertex at the point (h, k). Being able to seamlessly convert between these forms allows us to leverage the strengths of each representation. The conversion process itself reinforces fundamental algebraic skills, such as expanding squared terms, distributing coefficients, and combining constants. It also fosters a deeper understanding of how each term contributes to the overall function. Furthermore, this mastery extends beyond theoretical exercises. It empowers us to model and solve real-world problems involving quadratic relationships, from projectile motion to optimization scenarios. Therefore, dedicating time and effort to mastering these conversions is an investment in a broader understanding of algebra and its applications.