Unlocking The Secrets Of Parabolas: A Step-by-Step Guide

Hey guys! Ever wondered how to unlock the secrets hidden within a parabola? Let's dive deep into the world of quadratic functions, specifically the equation y = -x² + 6x - 2. We'll be on a quest to find some key features: the vertex (the peak or valley), the points where the parabola crosses the x-axis (x-intercepts), and where it kisses the y-axis (y-intercept). And, of course, we'll wrap things up by sketching the beautiful curve itself – the parabola! Buckle up, because we're about to embark on an awesome mathematical adventure.

Finding the Vertex: The Heart of the Parabola

Alright, let's start with the vertex. Think of the vertex as the most important point of the parabola – it's either the highest point (if the parabola opens downwards, like ours) or the lowest point (if it opens upwards). Finding the vertex is super important. To find this special spot, we can use a handy formula derived from the standard form of a quadratic equation. The standard form is y = ax² + bx + c. In our case, a = -1, b = 6, and c = -2. Note that the value of a determines whether the parabola opens upwards or downwards. If a is negative, like in our case, the parabola opens downwards, meaning we'll have a maximum point (a peak) as our vertex. If a is positive, the parabola opens upwards, leading to a minimum point (a valley).

To find the x-coordinate of the vertex (let's call it x_v), we use the formula: x_v = -b / 2a. Plugging in our values, we get x_v = -6 / (2 * -1) = 3. So, the x-coordinate of our vertex is 3. Now, to find the y-coordinate of the vertex (let's call it y_v), we substitute the x-coordinate we just found (which is 3) back into our original equation: y = -x² + 6x - 2. This gives us y_v = -(3)² + 6(3) - 2 = -9 + 18 - 2 = 7. Therefore, the vertex of our parabola is the point (3, 7). This means the peak of our parabola is at the coordinates (3, 7). We've successfully found the heart of our parabola! Isn't math cool? We can also find the vertex using a method called completing the square. It involves manipulating the quadratic equation into a form that easily reveals the vertex. The completed square form is y = a(x-h)² + k, where (h, k) is the vertex of the parabola. We can rewrite the equation y = -x² + 6x - 2 as follows: Factor out the negative sign: y = -(x² - 6x) - 2. Complete the square inside the parenthesis by adding and subtracting (6/2)² = 9. So, it becomes y = -(x² - 6x + 9 - 9) - 2. Now, rewrite as y = -((x - 3)² - 9) - 2. Distribute the negative sign to become y = -(x - 3)² + 9 - 2. Simplify it to y = -(x - 3)² + 7. Thus the vertex becomes (3,7).

Uncovering the Y-Intercept: Where the Parabola Meets the Y-Axis

Next up, let's find the y-intercept. This is the point where the parabola crosses the y-axis. The y-intercept always occurs when x = 0. So, to find it, we simply plug x = 0 into our equation: y = -x² + 6x - 2. This gives us y = -(0)² + 6(0) - 2 = -2. Therefore, the y-intercept is the point (0, -2). This tells us that our parabola crosses the y-axis at the point (0, -2). It's always a straightforward calculation! Just remember, the y-intercept is simply the value of 'c' in the standard form y = ax² + bx + c.

Hunting Down the X-Intercepts: The Roots of the Parabola

Now, let's hunt for the x-intercepts. These are the points where the parabola crosses the x-axis. At these points, y = 0. So, we need to solve the equation 0 = -x² + 6x - 2. This is a quadratic equation, and there are a couple of ways to solve it. We could try to factor it, but in this case, factoring isn't straightforward. Therefore, we will use the quadratic formula. The quadratic formula is x = (-b ± √(b² - 4ac)) / 2a. Applying this to our equation (a = -1, b = 6, c = -2), we get x = (-6 ± √(6² - 4 * -1 * -2)) / (2 * -1). Simplifying this, we get x = (-6 ± √(36 - 8)) / -2 = (-6 ± √28) / -2. Now, we can simplify √28 to 2√7. Hence, x = (-6 ± 2√7) / -2. Dividing both the terms by -2, we get, x = 3 ± √7. So, we have two x-intercepts: x = 3 + √7 and x = 3 - √7. Using a calculator, we can approximate these values: x ≈ 5.65 and x ≈ 0.35. This means our parabola crosses the x-axis at approximately (5.65, 0) and (0.35, 0). It's also possible that a parabola doesn't have any real x-intercepts. This happens when the discriminant (b² - 4ac) in the quadratic formula is negative. In this case, the square root of a negative number would be involved, resulting in complex (non-real) roots and, therefore, no x-intercepts on the standard graph.

Crafting the Parabola: Putting It All Together

Alright, awesome! We've found the vertex (3, 7), the y-intercept (0, -2), and the x-intercepts (approximately 5.65 and 0.35). Now, let's put it all together and sketch our parabola. First, plot the vertex (3, 7). Since the coefficient 'a' is negative, we know the parabola opens downwards, forming a