Find The Y-Intercept: F(x) = -x² - 2x + 3

Hey everyone! Today, we're diving into the world of quadratic functions, specifically how to find the y-intercepts of the function given by the equation f(x) = -x^2 - 2x + 3. This is a fundamental concept in algebra and understanding it can unlock a lot of insights about the behavior of parabolas, the graphs of quadratic functions. So, let's get started and break it down step by step!

Understanding Quadratic Functions and $y$-intercepts

Before we jump into the calculations, let's make sure we're all on the same page about what quadratic functions and $y$-intercepts actually are. A quadratic function is a polynomial function of degree two, meaning the highest power of the variable x is 2. The general form of a quadratic function is f(x) = ax^2 + 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 U-shaped curve that can open upwards or downwards depending on the sign of the coefficient a. In our case, f(x) = -x^2 - 2x + 3, a is -1, b is -2, and c is 3.

The y-intercept, on the other hand, is the point where the graph of the function intersects the y-axis. Remember that the y-axis is the vertical line where x is equal to 0. So, to find the $y$-intercept, we need to find the value of f(x) when x is 0. This is because any point on the y-axis has an x-coordinate of 0. Graphically, the $y$-intercept is where the parabola crosses the vertical y-axis. It's a crucial point that gives us a quick understanding of where the graph starts (or ends) its vertical journey. Think of it as the function's starting point when viewed from the y-axis perspective. This point is extremely useful when sketching the graph of the quadratic function, as it provides one of the key points to anchor the parabola. The $y$-intercept also has practical applications in various real-world scenarios modeled by quadratic functions, such as projectile motion or the shape of suspension cables. Understanding the $y$-intercept, therefore, isn't just about algebraic manipulation; it's about grasping a fundamental aspect of the function's behavior and its relevance in different contexts.

The Easy Way to Find the $y$-intercept: Setting $x = 0$

Okay, guys, the trick to finding the $y$-intercept is super straightforward. Since the $y$-intercept is the point where the graph crosses the y-axis, and we know that every point on the y-axis has an x-coordinate of 0, all we need to do is plug in x = 0 into our function and solve for f(x). This is because when x = 0, f(x) will give us the y-coordinate of the point where the graph intersects the y-axis. So, in simpler terms, we're finding the value of the function when x is zero, which directly tells us where the graph hits the y-axis.

For our function, f(x) = -x^2 - 2x + 3, we substitute x with 0:

f(0) = -(0)^2 - 2(0) + 3

Notice how everywhere we see an x, we replace it with a 0. Now, we just simplify the expression. Remember the order of operations (PEMDAS/BODMAS): Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). So first, we deal with the exponents:

f(0) = -0 - 2(0) + 3

Next, we perform the multiplication:

f(0) = 0 - 0 + 3

Finally, we do the addition and subtraction:

f(0) = 3

And there you have it! We found that f(0) = 3. This means that when x is 0, y (or f(x)) is 3. Therefore, the $y$-intercept is the point (0, 3). It's that simple! By plugging in 0 for x, we've directly calculated the y-coordinate where the parabola intersects the y-axis. This method works for any function, not just quadratic ones, making it a powerful tool in your mathematical toolkit. The key takeaway here is that the $y$-intercept is always found where x equals zero, simplifying the process to a straightforward substitution and evaluation.

The $y$-intercept in Our Example: $(0, 3)$

So, as we just calculated, the $y$-intercept of the quadratic function f(x) = -x^2 - 2x + 3 is the point (0, 3). This point is where the parabola crosses the vertical y-axis on the coordinate plane. Let's break down what this means visually and practically.

Visually, if you were to graph this quadratic function, you would see a parabola that opens downwards (because the coefficient of the x^2 term is negative, which is -1 in this case). The point (0, 3) would be one of the key points on that parabola, specifically the point where the curve intersects the y-axis. This intersection gives us a clear anchor point for sketching the graph and understanding its vertical positioning. The parabola will curve around this point, either going above and then down, or just curving downwards from this starting point on the y-axis.

Practically, the $y$-intercept can tell us a lot about the function's behavior in real-world scenarios. For example, if this function represented the height of a projectile over time, the $y$-intercept (0, 3) would represent the initial height of the projectile at time t = 0. Imagine throwing a ball; the $y$-intercept would be the height of the ball at the moment it leaves your hand. Similarly, in business applications, if the function represented profit over time, the $y$-intercept could indicate the initial profit (or loss) at the start of the period. This makes the $y$-intercept a crucial piece of information for interpreting the function's meaning in various contexts. It provides a starting point or a baseline, helping us understand the initial conditions or the state of the system at the beginning of the modeled scenario. In essence, knowing the $y$-intercept gives us a tangible sense of the function's practical implications and its relevance to real-world problems.

Why the Constant Term Matters: Connecting '$c$' to the $y$-intercept

Here's a cool connection to make: Did you notice that the $y$-coordinate of the $y$-intercept, which we found to be 3, is the same as the constant term c in our quadratic function f(x) = -x^2 - 2x + 3? This isn't just a coincidence, guys; it's a fundamental property of quadratic functions (and polynomial functions in general!). The constant term c in the quadratic equation f(x) = ax^2 + bx + c always represents the $y$-intercept of the parabola.

Let's think about why this is the case. When we set x = 0 to find the $y$-intercept, the terms involving x (that is, ax^2 and bx) become zero. This is because any number multiplied by zero is zero. So, the equation simplifies to f(0) = a(0)^2 + b(0) + c = 0 + 0 + c = c. This shows us that the value of the function when x is 0 is simply c, which is the $y$-coordinate of the $y$-intercept.

This connection is super useful because it gives us a quick way to identify the $y$-intercept without doing any calculations. Just look at the equation, and the constant term is your $y$-intercept! For example, if you have the equation f(x) = 2x^2 - 5x + 7, you immediately know that the $y$-intercept is (0, 7). This shortcut can save you time on tests and make understanding graphs much easier. Furthermore, recognizing this relationship reinforces the importance of the standard form of a quadratic equation. It highlights how each coefficient plays a specific role in determining the parabola's characteristics. The coefficient a determines the direction and width of the opening, b influences the axis of symmetry, and, as we've discussed, c directly reveals the $y$-intercept. Understanding this interplay allows for a more intuitive grasp of quadratic functions and their graphical representations.

Practice Makes Perfect: Examples and Exercises

Alright, guys, now that we've covered the theory and the method, let's put our knowledge into practice with some examples and exercises. Working through these will help solidify your understanding and make finding $y$-intercepts a breeze!

Example 1:

Find the $y$-intercept of the quadratic function g(x) = 3x^2 + 4x - 5.

Solution:

Using our shortcut, we know that the $y$-intercept is the constant term, which is -5. Therefore, the $y$-intercept is (0, -5). Easy peasy, right?

Example 2:

What is the $y$-intercept of the function h(x) = -2x^2 + 7x?

Solution:

Notice that this function doesn't have a constant term explicitly written. This means the constant term is 0. So, the $y$-intercept is (0, 0). This tells us the parabola passes through the origin.

Exercises:

1. Find the $y$-intercept of f(x) = x^2 - 6x + 8.
2. Determine the $y$-intercept of g(x) = -4x^2 + 9.
3. What is the $y$-intercept of h(x) = 5x^2 - 3x?

Try working through these exercises on your own. Remember to set x = 0 or use the shortcut of identifying the constant term. The more you practice, the more comfortable you'll become with finding $y$-intercepts. And if you get stuck, don't worry! Review the steps we've discussed, and remember that the $y$-intercept is simply the point where the graph crosses the y-axis, which occurs when x is zero. The process of solving these exercises not only reinforces the method but also sharpens your problem-solving skills. It helps you recognize patterns and apply the concept in various forms, ensuring a deeper understanding of the topic. So, grab a pen and paper, give these a try, and watch your confidence in finding $y$-intercepts soar!

Conclusion: Mastering the $y$-intercept

Great job, guys! You've now learned how to find the $y$-intercept of a quadratic function. Remember, the $y$-intercept is the point where the graph crosses the y-axis, and you can find it by setting x = 0 in the equation. Even better, you can simply identify the constant term in the quadratic equation, as it directly gives you the $y$-coordinate of the $y$-intercept. This skill is super valuable for understanding and graphing quadratic functions.

Finding the $y$-intercept is not just a mathematical exercise; it's a fundamental skill that unlocks deeper insights into the behavior of quadratic functions and their applications. Whether you're sketching a parabola, analyzing projectile motion, or modeling business profits, the $y$-intercept provides a crucial reference point. It gives you a starting value, a baseline from which to understand the function's progression. The ability to quickly determine the $y$-intercept, either by setting x to zero or by recognizing the constant term, empowers you to grasp the essence of the function at a glance. Moreover, mastering this concept builds a solid foundation for tackling more complex topics in algebra and calculus. It's a building block that supports your understanding of graphs, transformations, and the relationships between algebraic expressions and their visual representations. So, keep practicing, keep exploring, and you'll find that the $y$-intercept is a powerful tool in your mathematical journey.