Analyzing The Quadratic Function F(x) = 2x² + 8x - 24 Graph Direction Intercepts And Vertex

Hey guys! Let's dive into the fascinating world of quadratic functions, where curves reign supreme and parabolas dance across the coordinate plane. Today, we're going to dissect the function f(x) = 2x² + 8x - 24, uncovering its secrets and revealing its graphical personality. We'll explore its direction, pinpoint its intercepts, and locate its vertex, giving you a comprehensive understanding of this quadratic gem.

Decoding the Direction: Is It an Upward or Downward Curve?

When we talk about the direction of a quadratic function's graph, we're essentially asking: does it open upwards like a smile or downwards like a frown? The key to unlocking this lies in the coefficient of the x² term. In our case, f(x) = 2x² + 8x - 24, the coefficient is 2. This is a positive number, and that's our golden ticket! A positive coefficient means the parabola opens upwards. Think of it like this: a positive outlook on life makes you smile, and a positive coefficient makes the parabola smile too!

But why is this the case, you might ask? Well, as x moves further away from the vertex (either to the left or the right), the x² term dominates the function's behavior. Since x² is always positive (or zero), multiplying it by a positive coefficient (like our 2) ensures that the function's values become increasingly positive as you move away from the vertex. This creates the upward-opening shape.

Imagine plotting points for very large positive and negative values of x. For instance, if x = 100, f(x) becomes incredibly large and positive due to the 2x² term overpowering the other terms. Similarly, if x = -100, f(x) also becomes large and positive. This behavior dictates the upward trajectory of the parabola's arms, confirming its upward-opening nature. So, remember, a positive coefficient for the x² term always spells an upward-opening parabola, and in our case, f(x) = 2x² + 8x - 24 proudly opens upwards, ready to reveal its other secrets.

Unveiling the Y-Intercept: Where the Parabola Crosses the Vertical Axis

The y-intercept is a crucial point on any graph, marking where the function intersects the vertical y-axis. It's like a friendly handshake between the function and the y-axis, giving us a valuable piece of information about the function's behavior. To find the y-intercept, we simply set x = 0 in our function, f(x) = 2x² + 8x - 24. This is because any point on the y-axis has an x-coordinate of 0.

Let's plug in x = 0 and see what happens: f(0) = 2(0)² + 8(0) - 24. This simplifies beautifully: f(0) = 0 + 0 - 24. Therefore, f(0) = -24. Ah-ha! We've found our y-intercept. It's the point where the graph crosses the y-axis, and it occurs at y = -24. This tells us that the parabola dips down quite a bit before making its upward turn. The y-intercept gives us a starting point, a landmark on the y-axis that helps us visualize the parabola's position. It's like knowing the starting line of a race – it gives you context for the rest of the journey.

Think of the y-intercept as the parabola's initial greeting as it enters the coordinate plane. It's the first point we encounter as we move along the y-axis, and it provides a direct connection to the constant term in the quadratic equation. In general, for a quadratic function in the form f(x) = ax² + bx + c, the y-intercept is simply the constant term, 'c'. This is because when x = 0, the ax² and bx terms vanish, leaving us with just 'c'. So, in our case, the y-intercept of -24 is immediately apparent from the equation f(x) = 2x² + 8x - 24. Knowing this shortcut can save you time and effort when identifying the y-intercept of any quadratic function.

Discovering the X-Intercepts: Where the Parabola Meets the Horizontal Axis

Now, let's hunt for the x-intercepts, those elusive points where the parabola gracefully crosses the horizontal x-axis. These points are also known as the roots or zeros of the function, and they hold significant information about the parabola's behavior and its solutions. To find the x-intercepts, we need to solve the equation f(x) = 0, meaning we set 2x² + 8x - 24 = 0.

There are a few ways we can tackle this quadratic equation. One popular method is factoring. Let's see if we can factor the expression 2x² + 8x - 24. First, notice that all the coefficients are divisible by 2, so we can factor out a 2: 2(x² + 4x - 12) = 0. Now we need to factor the quadratic expression inside the parentheses. We're looking for two numbers that multiply to -12 and add up to 4. Those numbers are 6 and -2. So, we can factor the expression as 2(x + 6)(x - 2) = 0.

To find the x-intercepts, we set each factor equal to zero: x + 6 = 0 or x - 2 = 0. Solving for x, we get x = -6 and x = 2. These are our x-intercepts! The parabola crosses the x-axis at x = -6 and x = 2. These points provide us with crucial anchors on the x-axis, helping us sketch the parabola's shape and position more accurately.

Another way to find the x-intercepts is by using the quadratic formula. This formula is a powerful tool that can solve any quadratic equation, regardless of whether it can be factored easily. The quadratic formula is: x = [-b ± √(b² - 4ac)] / 2a, where a, b, and c are the coefficients of the quadratic equation ax² + bx + c = 0. In our case, a = 2, b = 8, and c = -24. Plugging these values into the quadratic formula, we get: x = [-8 ± √(8² - 4 * 2 * -24)] / (2 * 2). Simplifying this expression will lead us to the same x-intercepts: x = -6 and x = 2. So, whether you prefer factoring or the quadratic formula, the x-intercepts remain the same, marking the parabola's encounters with the x-axis.

Locating the Vertex: The Turning Point of the Parabola

The vertex is the most special point on a parabola – it's the turning point, the place where the parabola changes direction. For an upward-opening parabola like ours, the vertex is the minimum point. It's the bottom of the curve, the place where the function reaches its lowest value. Finding the vertex is like finding the heart of the parabola, revealing its central characteristic.

There are a couple of ways to find the vertex. One way is to use the vertex formula. For a quadratic function in the form f(x) = ax² + bx + c, the x-coordinate of the vertex is given by x = -b / 2a. In our case, a = 2 and b = 8, so the x-coordinate of the vertex is x = -8 / (2 * 2) = -8 / 4 = -2. To find the y-coordinate of the vertex, we plug this x-value back into our function: f(-2) = 2(-2)² + 8(-2) - 24 = 2(4) - 16 - 24 = 8 - 16 - 24 = -32. So, the vertex is located at the point (-2, -32).

Another way to find the vertex is by completing the square. This method involves rewriting the quadratic function in vertex form, which is f(x) = a(x - h)² + k, where (h, k) is the vertex. Let's complete the square for our function, f(x) = 2x² + 8x - 24. First, factor out the coefficient of the x² term from the first two terms: f(x) = 2(x² + 4x) - 24. Now, we need to add and subtract a value inside the parentheses to create a perfect square trinomial. Take half of the coefficient of the x term (which is 4), square it (which is 4), and add and subtract it inside the parentheses: f(x) = 2(x² + 4x + 4 - 4) - 24. Now, rewrite the perfect square trinomial: f(x) = 2((x + 2)² - 4) - 24. Distribute the 2: f(x) = 2(x + 2)² - 8 - 24. Finally, combine the constant terms: f(x) = 2(x + 2)² - 32. This is the vertex form of the equation, and we can see that the vertex is at (-2, -32), confirming our previous result.

The vertex (-2, -32) is the lowest point on our parabola. It's a critical point that helps us understand the parabola's symmetry and its range (the set of all possible y-values). Since the parabola opens upwards, the y-coordinate of the vertex (-32) is the minimum value of the function. The vertex, in essence, is the anchor point from which the parabola extends its graceful curves, either upwards or downwards, defining its overall shape and position in the coordinate plane.

Putting It All Together: A Complete Picture of f(x) = 2x² + 8x - 24

Wow, we've uncovered a lot about our quadratic function, f(x) = 2x² + 8x - 24! We determined that the graph opens upwards because the coefficient of the x² term is positive. We found the y-intercept to be at y = -24, marking where the parabola crosses the vertical axis. We discovered the x-intercepts at x = -6 and x = 2, the points where the parabola intersects the horizontal axis. And finally, we located the vertex at (-2, -32), the turning point and minimum of the function.

With this information, we can confidently sketch the graph of f(x) = 2x² + 8x - 24. Imagine a U-shaped curve opening upwards, dipping down to its vertex at (-2, -32), crossing the x-axis at -6 and 2, and intersecting the y-axis at -24. This is the visual representation of our algebraic exploration, bringing the function to life on the coordinate plane.

Understanding the direction, intercepts, and vertex of a quadratic function allows us to not only sketch its graph but also to solve real-world problems involving parabolic shapes, such as projectile motion, bridge design, and optimization scenarios. Quadratic functions are powerful tools, and by mastering their characteristics, we unlock a deeper understanding of the mathematical world around us. So, keep exploring, keep questioning, and keep uncovering the secrets hidden within the elegant curves of quadratic functions!