Quadratic Function F(x) = -2x² + 2x + 1 Analysis And Graphing

Hey guys! Today, we're diving deep into the world of quadratic functions, focusing specifically on the function f(x) = -2x² + 2x + 1. We'll be tackling this problem step-by-step, starting with completing the square, then identifying the maximum or minimum point, and finally sketching the graph. So, buckle up and let's get started!

(i) Completing the Square for f(x) = -2x² + 2x + 1

The first task at hand is to express our quadratic function, f(x) = -2x² + 2x + 1, in the vertex form, which is f(x) = a(x + p)² + q. This form is super helpful because it directly reveals the vertex (or turning point) of the parabola. The process of getting there is called completing the square, and it's a neat trick that allows us to rewrite the quadratic expression in a more insightful way. Think of it as rearranging the furniture in a room to make it more functional and visually appealing!

So, how do we actually complete the square? Let's break it down. The initial step involves factoring out the coefficient of the x² term from the first two terms of the quadratic. In our case, that coefficient is -2. By factoring out -2 from -2x² + 2x, we get -2(x² - x). This maneuver is crucial because it sets the stage for creating a perfect square trinomial inside the parentheses. A perfect square trinomial, as the name suggests, is a trinomial that can be factored into the square of a binomial. For example, x² + 2x + 1 is a perfect square trinomial because it can be written as (x + 1)². Recognizing and creating these trinomials is the heart of completing the square.

Now, we focus on the expression inside the parentheses, x² - x. To turn this into a perfect square trinomial, we need to add and subtract a specific value. This value is determined by taking half of the coefficient of the x term (which is -1 in this case), squaring it, and then adding and subtracting the result. Half of -1 is -1/2, and squaring it gives us 1/4. So, we add and subtract 1/4 inside the parentheses. This might seem like we're just making things more complicated, but trust me, it's a clever way to rewrite the expression without changing its value. We're essentially adding zero in a disguised form!

By adding and subtracting 1/4, our expression inside the parentheses becomes x² - x + 1/4 - 1/4. The first three terms, x² - x + 1/4, now form a perfect square trinomial, which can be factored as (x - 1/2)². The remaining term, -1/4, is just a constant that we'll deal with later. So, we've successfully transformed the expression inside the parentheses into a squared term plus a constant, which is exactly what we wanted.

Next, we substitute this back into our original expression. We had -2(x² - x), which now becomes -2((x - 1/2)² - 1/4). We then distribute the -2 to both terms inside the parentheses. Multiplying -2 by (x - 1/2)² gives us -2(x - 1/2)², and multiplying -2 by -1/4 gives us +1/2. So, our expression now looks like -2(x - 1/2)² + 1/2. But remember, we still have the original +1 in our function f(x) = -2x² + 2x + 1. We need to add this +1 to our current expression.

Finally, we add the +1 to the +1/2 we obtained earlier. This gives us 1/2 + 1 = 3/2. Therefore, our quadratic function f(x) = -2x² + 2x + 1, when expressed in the vertex form by completing the square, becomes f(x) = -2(x - 1/2)² + 3/2. This is a significant result! We've successfully rewritten the function in a form that readily reveals its key characteristics. The values of a, p, and q in the vertex form f(x) = a(x + p)² + q are a = -2, p = -1/2, and q = 3/2. Understanding these values is crucial for the next part of our exploration.

(ii) Stating the Maximum or Minimum Point of the Function

Now that we've successfully expressed our quadratic function in vertex form, f(x) = -2(x - 1/2)² + 3/2, we're in a prime position to identify its maximum or minimum point. This point, also known as the vertex of the parabola, is the turning point of the graph. It's where the function transitions from increasing to decreasing or vice versa. Finding this point is like locating the peak of a mountain or the bottom of a valley – it gives us a clear sense of the function's behavior.

The vertex form of a quadratic function, f(x) = a(x + p)² + q, provides us with a direct way to determine the vertex. The vertex is given by the coordinates (-p, q). Notice the sign change for the x-coordinate! In our case, we have f(x) = -2(x - 1/2)² + 3/2, so p = -1/2 and q = 3/2. Therefore, the x-coordinate of the vertex is -(-1/2) = 1/2, and the y-coordinate is 3/2. This means the vertex of our parabola is at the point (1/2, 3/2).

But how do we know whether this point represents a maximum or a minimum? That's where the coefficient 'a' in the vertex form comes into play. The sign of 'a' tells us about the concavity of the parabola. If 'a' is positive, the parabola opens upwards, forming a U-shape, and the vertex represents the minimum point. Think of it like a smiley face – the lowest point is the minimum. Conversely, if 'a' is negative, the parabola opens downwards, forming an inverted U-shape, and the vertex represents the maximum point. This is like a frowny face – the highest point is the maximum.

In our function, f(x) = -2(x - 1/2)² + 3/2, the coefficient 'a' is -2, which is negative. This tells us that the parabola opens downwards, and therefore, the vertex (1/2, 3/2) represents the maximum point of the function. This means that the function reaches its highest value at x = 1/2, and that maximum value is 3/2. Understanding this is crucial for sketching the graph and for any applications of this quadratic function. For instance, if this function represented the height of a projectile, we would know that the projectile reaches its maximum height of 3/2 units at a time of x = 1/2 units.

So, to summarize, by analyzing the vertex form of the quadratic function, we've not only pinpointed the vertex but also determined whether it's a maximum or minimum. This is a powerful illustration of how rewriting an equation can reveal hidden information and provide deeper insights into the function's behavior. Remember, the sign of 'a' is your guide to concavity, and the coordinates (-p, q) directly give you the vertex. Keep these principles in mind, and you'll be able to conquer any quadratic function that comes your way!

(iii) Sketching the Graph of f(x) = -2x² + 2x + 1

Alright, guys, let's bring this all together and sketch the graph of our quadratic function, f(x) = -2x² + 2x + 1. We've already done the heavy lifting by completing the square and finding the maximum point. Now, it's time to visualize what this function looks like on a coordinate plane. Sketching the graph helps us to solidify our understanding of the function's behavior and to see the relationship between the equation and its visual representation.

We know that our function is a parabola because it's a quadratic function. Parabolas are U-shaped curves, and we've already determined that our parabola opens downwards because the coefficient of the x² term is negative (-2). We also know that the vertex, which is the maximum point in this case, is located at (1/2, 3/2). This gives us a crucial starting point for our sketch.

To sketch the graph, we need a few more points besides the vertex. A good strategy is to find the x-intercepts and the y-intercept. The x-intercepts are the points where the graph crosses the x-axis, which means f(x) = 0. To find them, we need to solve the equation -2x² + 2x + 1 = 0. This is a quadratic equation, and we can solve it using the quadratic formula.

The quadratic formula is a powerful tool that provides the solutions (or roots) of any quadratic equation in the form ax² + bx + c = 0. The formula is x = (-b ± √(b² - 4ac)) / (2a). In our case, a = -2, b = 2, and c = 1. Plugging these values into the quadratic formula, we get:

x = (-2 ± √(2² - 4(-2)(1))) / (2(-2))

x = (-2 ± √(4 + 8)) / (-4)

x = (-2 ± √12) / (-4)

x = (-2 ± 2√3) / (-4)

Simplifying this, we get two x-intercepts:

x₁ = (1 - √3) / 2 ≈ -0.37

x₂ = (1 + √3) / 2 ≈ 1.37

So, our parabola crosses the x-axis at approximately x = -0.37 and x = 1.37. These points give us a good sense of the parabola's width and position relative to the x-axis.

Next, let's find the y-intercept. The y-intercept is the point where the graph crosses the y-axis, which means x = 0. To find it, we simply substitute x = 0 into our function: f(0) = -2(0)² + 2(0) + 1 = 1. So, the y-intercept is at the point (0, 1).

Now we have enough information to sketch the graph! We have the vertex at (1/2, 3/2), the x-intercepts at approximately (-0.37, 0) and (1.37, 0), and the y-intercept at (0, 1). We know the parabola opens downwards, so we can draw a smooth curve that passes through these points, with the vertex being the highest point on the graph. Imagine connecting the dots to create the familiar U-shape, but inverted.

When sketching, remember to label the key points, such as the vertex and intercepts. This helps to clearly communicate the important features of the graph. Also, consider the symmetry of the parabola. Parabolas are symmetrical about the vertical line that passes through the vertex. This means that the part of the parabola to the left of the vertex is a mirror image of the part to the right. This symmetry can be a helpful guide when sketching the curve.

By sketching the graph, we've visually represented the behavior of our quadratic function. We can see the maximum point, the points where the graph intersects the axes, and the overall shape of the parabola. This visual representation complements our algebraic analysis and provides a more complete understanding of the function. Graphing is a powerful tool in mathematics, and mastering it will help you to tackle a wide range of problems.

In conclusion, we've successfully navigated the world of quadratic functions, from completing the square to sketching the graph. We've seen how different forms of the equation reveal different aspects of the function, and how a visual representation can solidify our understanding. Keep practicing these skills, and you'll become a quadratic function pro in no time!