Unlocking The Secrets Of Quadratic Functions: A Detailed Guide

Hey everyone! Today, we're diving deep into the fascinating world of quadratic functions. We'll be taking a close look at a specific example: f(x) = 2x^2 - 7x - 30. Our mission? To uncover its secrets! We'll find its vertex, pinpoint those x-intercepts, and snag that y-intercept. Ready to get started?

Unveiling the Vertex: The Heart of the Parabola

Let's start with the vertex, the most crucial point of our quadratic function. Think of the vertex as the heart of the parabola, the point where it changes direction. It's either the lowest point (if the parabola opens upwards) or the highest point (if it opens downwards). For the quadratic function f(x) = 2x^2 - 7x - 30, finding the vertex is our first priority. We can do this using a handy formula. The x-coordinate of the vertex can be found using the formula: x = -b / 2a. Where 'a' and 'b' come from the standard form of a quadratic equation ax^2 + bx + c. In our case, a = 2, and b = -7. Let's plug those values in: x = -(-7) / (2 * 2) = 7/4. So, the x-coordinate of our vertex is 7/4, or 1.75.

Now, to find the y-coordinate of the vertex, we need to substitute this x-value back into the original function: f(7/4) = 2*(7/4)^2 - 7*(7/4) - 30. Doing the calculations, we get f(7/4) = 2*(49/16) - 49/4 - 30. This simplifies to f(7/4) = 49/8 - 98/8 - 240/8 = -289/8. So, the y-coordinate of the vertex is -289/8, which is about -36.125. Therefore, the vertex of the parabola is located at the point (7/4, -289/8), or (1.75, -36.125). Because the coefficient 'a' is positive (a = 2), we know this parabola opens upwards, and the vertex represents the minimum point on the curve. This is super important to understand because it tells us the function's lowest value.

So, to recap, finding the vertex involves calculating the x-coordinate using the formula -b/2a and then plugging that value back into the original equation to solve for the y-coordinate. This process helps us identify the key turning point of our quadratic function, providing valuable information about its behavior and properties. Also, to reiterate, it's very important to note that the vertex provides the minimum or maximum value of the function, depending on whether the parabola opens up or down. Got it? Awesome!

Pinpointing the x-intercepts: Where the Curve Meets the x-axis

Next up, let's find the x-intercepts. These are the points where our parabola crosses the x-axis. At these points, the value of the function f(x) is zero. To find them, we need to solve the equation f(x) = 0, which means 2x^2 - 7x - 30 = 0. There are several ways to solve a quadratic equation. We can use factoring, completing the square, or the quadratic formula. Let's start by attempting to factor the quadratic expression. We need to find two numbers that multiply to give us (2 * -30) = -60 and add up to -7. After some thinking, we find that the numbers are -12 and 5. We can rewrite the middle term (-7x) as -12x + 5x. Then, we can rewrite the equation as follows: 2x^2 - 12x + 5x - 30 = 0. Now, let’s factor by grouping:

• Factor out 2x from the first two terms: 2x(x - 6).
• Factor out 5 from the last two terms: 5(x - 6).

Now, the equation becomes: 2x(x - 6) + 5(x - 6) = 0. Notice that we have a common factor of (x - 6). We can factor this out to get (x - 6)(2x + 5) = 0. So, we set each factor equal to zero and solve for x. For (x - 6) = 0, we get x = 6. For (2x + 5) = 0, we get 2x = -5, which simplifies to x = -5/2, or -2.5. Therefore, the x-intercepts of the function are x = 6 and x = -5/2. These are the points where the parabola intersects the x-axis.

These x-intercepts are also known as the roots or zeros of the function, and they represent the solutions to the quadratic equation. Graphically, the x-intercepts tell us where the function's output (y-value) is zero. Finding the x-intercepts is important as it helps us understand the behavior of the quadratic function and where it crosses the x-axis, which is often crucial when we analyze real-world scenarios that can be modeled by a quadratic equation. Remember, if we couldn't factor the equation, we could have used the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. This formula always works, even when factoring is difficult or impossible. It's a handy tool to have in your mathematical toolkit! We'll cover it later if we need it.

The y-intercept: Meeting the y-axis

Finally, let's find the y-intercept. This is the point where the parabola crosses the y-axis. At this point, the value of x is zero. To find the y-intercept, we simply substitute x = 0 into the function: f(0) = 2(0)^2 - 7(0) - 30. This simplifies to f(0) = -30. Therefore, the y-intercept is at the point (0, -30). This is the point where the parabola intersects the y-axis. The y-intercept tells us the value of the function when x is zero. It's the starting point of our parabola. In this case, when x is zero, the value of the function is -30. This means the parabola crosses the y-axis at the point (0, -30).

Understanding the y-intercept gives us another key point on the graph. It helps visualize where the parabola begins its journey. In many real-world applications, the y-intercept can represent an initial value or starting point, depending on the scenario the quadratic function is modeling. So, finding the y-intercept is as simple as plugging in zero for x and solving for f(x).

Summary: Putting It All Together

Alright, let's wrap things up! For the quadratic function f(x) = 2x^2 - 7x - 30, here's what we found:

• Vertex: (7/4, -289/8) or (1.75, -36.125)
• x-intercepts: x = 6 and x = -5/2
• y-intercept: y = -30

We successfully identified the vertex, which is the minimum point. The x-intercepts show us where the graph crosses the x-axis, and the y-intercept reveals where it crosses the y-axis. By knowing these key points, we have a clear understanding of the shape and position of the parabola on the coordinate plane. These parameters (vertex, x-intercepts, and y-intercept) give us a comprehensive view of the quadratic function's behavior. Keep in mind that understanding these elements allows you to graph the function, solve related problems, and interpret its real-world implications.

Learning to analyze quadratic functions allows you to solve real-world problems. Whether it's the trajectory of a ball, the shape of a bridge, or optimization problems in business, quadratic functions are essential tools. Practice is key, so keep working through examples and you'll become a quadratic function master! Great job, everyone! And remember, math is a journey, and every step counts. Keep exploring, keep learning, and keep asking questions. You've got this!