Finding The Y-intercept Of F(x) = (x-6)(x-2) A Comprehensive Guide

In this comprehensive guide, we'll explore how to determine the y-intercept of a quadratic function, using the specific example of $f(x) = (x-6)(x-2)$. The y-intercept is a fundamental concept in understanding the behavior and graph of any function, and it represents the point where the function's graph intersects the y-axis. We'll break down the process step-by-step, ensuring you have a solid grasp of the underlying principles. We'll cover the definition of the y-intercept, its significance, and how to calculate it for quadratic functions. By the end of this discussion, you'll be equipped to find the y-intercept of any quadratic function given in factored form. We'll delve into the algebraic manipulations needed to arrive at the solution and highlight the connection between the function's equation and its graphical representation. Understanding the y-intercept is not only crucial for solving mathematical problems but also for visualizing and interpreting real-world scenarios modeled by quadratic functions. Let's embark on this mathematical journey together and unlock the secrets of the y-intercept.

Understanding the $y$-intercept

Before diving into the specifics of the given quadratic function, let's define what the y-intercept actually is. The y-intercept is the point where a graph intersects the y-axis. In other words, it's the point where the $x$-coordinate is zero. This is a crucial concept in coordinate geometry and function analysis. To find the y-intercept of any function, you simply need to set $x$ equal to zero and solve for $y$. This gives you the $y$-coordinate of the point where the graph crosses the y-axis. The y-intercept is often represented as the ordered pair $(0, y)$, where $y$ is the value of the function when $x$ is zero. Understanding the y-intercept provides valuable information about the function's behavior, especially its starting point on the vertical axis. It helps in sketching the graph of the function and in understanding its real-world implications. For example, in a quadratic function representing the trajectory of a projectile, the y-intercept might represent the initial height of the projectile. In the context of economics, the y-intercept could represent the initial cost or revenue. Therefore, grasping the concept of the y-intercept is essential for both mathematical problem-solving and practical applications.

Finding the $y$-intercept of $f(x) = (x-6)(x-2)$

Now, let's apply this concept to our specific quadratic function, $f(x) = (x-6)(x-2)$. To find the y-intercept, we need to determine the value of $f(x)$ when $x = 0$. This involves substituting $x = 0$ into the function and simplifying the expression. So, we have:

$f(0) = (0-6)(0-2)$

This simplifies to:

$f(0) = (-6)(-2)$

Multiplying these two negative numbers gives us:

$f(0) = 12$

Therefore, the y-intercept occurs when $y = 12$. This means the point where the graph of the function intersects the y-axis is $(0, 12)$. This point is a key feature of the parabola represented by the quadratic function. It tells us where the parabola starts its journey on the coordinate plane, specifically its vertical starting point. The y-intercept, along with other key points like the vertex and the x-intercepts, helps us sketch the complete graph of the quadratic function. Understanding this process is crucial for analyzing quadratic functions and their real-world applications. By substituting $x = 0$, we effectively isolate the constant term in the expanded form of the quadratic, which directly corresponds to the y-intercept. This method provides a straightforward and reliable way to find the y-intercept for any function.

Expanding the Quadratic Function (Optional)

While substituting $x=0$ into the factored form is the most direct method, we can also expand the quadratic function and see how the y-intercept is revealed in the standard form. Let's expand $f(x) = (x-6)(x-2)$:

$f(x) = x(x-2) - 6(x-2)$

$f(x) = x^2 - 2x - 6x + 12$

$f(x) = x^2 - 8x + 12$

Now, we have the quadratic function in the standard form, $f(x) = ax^2 + bx + c$, where $a = 1$, $b = -8$, and $c = 12$. Notice that the constant term, $c$, is equal to 12. This constant term always represents the y-intercept when the quadratic is in standard form. This is because when $x = 0$, the terms $ax^2$ and $bx$ both become zero, leaving only the constant term. So, the y-intercept is $(0, c)$, which in this case is $(0, 12)$. This method provides an alternative way to verify our previous result and reinforces the connection between the standard form of a quadratic function and its y-intercept. Expanding the quadratic can also be helpful in identifying other key features, such as the coefficients that determine the parabola's shape and direction. However, for the sole purpose of finding the y-intercept, substituting $x = 0$ into the factored form is generally more efficient.

The Correct Answer

Based on our calculations, the y-intercept of the quadratic function $f(x) = (x-6)(x-2)$ is $(0, 12)$. This corresponds to option B in the given choices. Options A, C, and D are incorrect. Option A, $(0, -6)$, might be a result of incorrectly multiplying the constants within the factors. Options C, $(-8, 0)$, and D, $(2, 0)$, represent x-intercepts (or roots) of the function, not the y-intercept. The x-intercepts are the points where the graph intersects the x-axis, meaning $f(x) = 0$. To find them, you would set the function equal to zero and solve for $x$, which is a different process than finding the y-intercept. Understanding the distinction between x-intercepts and the y-intercept is crucial for correctly interpreting the graph of a function. The y-intercept provides information about the function's value when $x$ is zero, while the x-intercepts provide information about where the function's value is zero. Therefore, carefully distinguishing between these concepts is essential for accurate problem-solving in mathematics.

Why Other Options are Incorrect

Let's briefly discuss why the other options are incorrect to solidify our understanding. Option A, $(0, -6)$, is incorrect because it seems to be a result of either adding the constants within the factors or misinterpreting the operation. When finding the y-intercept, we multiply $(0-6)$ and $(0-2)$, which results in $(-6)(-2) = 12$, not $-6$. This highlights the importance of carefully following the order of operations and correctly substituting the value of $x$. Option C, $(-8, 0)$, represents an x-intercept, not the y-intercept. X-intercepts are found by setting $f(x) = 0$ and solving for $x$. In this case, the x-intercepts are $x = 6$ and $x = 2$, as these values make each factor equal to zero. The point $(-8, 0)$ does not satisfy the equation $f(x) = 0$. Option D, $(2, 0)$, is also an x-intercept. While $x = 2$ is indeed a root of the function, the y-intercept is the point where the graph crosses the y-axis, not the x-axis. These incorrect options serve as a reminder to pay close attention to the definitions of y-intercepts and x-intercepts and to apply the correct procedures for finding them. By understanding why these options are wrong, we reinforce our grasp of the correct method and the underlying concepts.

Conclusion

In conclusion, the y-intercept of the quadratic function $f(x) = (x-6)(x-2)$ is $(0, 12)$. We arrived at this answer by substituting $x = 0$ into the function and simplifying the expression. We also explored how expanding the function into standard form reveals the y-intercept as the constant term. Furthermore, we discussed why the other options were incorrect, emphasizing the distinction between y-intercepts and x-intercepts. Understanding the y-intercept is a fundamental skill in algebra and calculus, and it plays a crucial role in analyzing and graphing functions. This concept allows us to quickly identify a key point on the graph and gain insights into the function's behavior. By mastering this technique, you'll be well-equipped to tackle a variety of problems involving quadratic functions and their applications. Remember, the y-intercept is where the graph crosses the y-axis, and it's found by setting $x = 0$. This simple yet powerful concept is a cornerstone of function analysis.