Determine The Range Of F(x) = 4 – 8x – X²

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Determine the Range of F(x) = 4 – 8x – x²

Hey guys! Today, we're going to dive into how to figure out the range of the function F(x) = 4 – 8x – x². This might sound a bit intimidating, but trust me, we'll break it down step by step so it's super easy to understand. We'll cover everything from the basics of functions and ranges to the nitty-gritty of completing the square to find the vertex. By the end of this, you'll be a pro at finding the range of quadratic functions!

Understanding Functions and Range

First things first, let's make sure we're all on the same page about what a function and its range actually are. Think of a function like a machine: you put something in (an input, usually 'x'), and it spits something out (an output, usually 'y' or F(x)'). The range is simply all the possible output values (y-values) that the function can produce. So, when we're asked to determine the range, we're essentially figuring out what the highest and lowest possible y-values are for the function.

For our specific function, F(x) = 4 – 8x – x², we have a quadratic function. You can tell because of the x² term. Quadratic functions always graph as parabolas, which are U-shaped curves (either opening upwards or downwards). This shape is super important because the vertex (the highest or lowest point on the parabola) plays a crucial role in determining the range. If the parabola opens upwards, the vertex is the minimum point, and the range includes all y-values greater than or equal to the y-coordinate of the vertex. If the parabola opens downwards, the vertex is the maximum point, and the range includes all y-values less than or equal to the y-coordinate of the vertex. So, the key to finding the range is often finding the vertex!

Identifying the Type of Quadratic Function

Before we jump into finding the range, we need to figure out if our parabola opens upwards or downwards. This is determined by the coefficient of the x² term. In our function, F(x) = 4 – 8x – x², the coefficient of x² is -1. Since it's negative, the parabola opens downwards. This means our parabola has a highest point (a maximum), and the range will include all y-values less than or equal to that maximum.

Think of it like a frown – if the coefficient of x² is negative, the parabola 'frowns' downwards. On the other hand, if the coefficient were positive, the parabola would 'smile' upwards. Knowing this direction is crucial because it tells us whether we're looking for a minimum or a maximum y-value. So, in our case, we know we're looking for the highest point on the parabola. This is a huge step in the right direction!

Completing the Square: The Key to Finding the Vertex

Okay, now for the fun part: finding the vertex! The most common method to do this for a quadratic function is called completing the square. Don't let the name scare you; it's just a clever algebraic trick. Completing the square transforms our quadratic function into a form that makes the vertex jump right out at us. The goal is to rewrite the function in the form F(x) = a(x – h)² + k, where (h, k) is the vertex of the parabola.

Let's walk through the steps for F(x) = 4 – 8x – x²:

  1. Rearrange the terms: First, let's rearrange the terms so the x² term comes first, followed by the x term, and then the constant: F(x) = -x² – 8x + 4.
  2. Factor out the coefficient of x²: Since the coefficient of x² is -1, we'll factor out a -1 from the first two terms: F(x) = -(x² + 8x) + 4. Remember, when you factor out a negative, it changes the signs inside the parentheses!
  3. Complete the square: Now comes the tricky part. We need to add and subtract a value inside the parentheses that will allow us to create a perfect square trinomial. A perfect square trinomial is a trinomial that can be factored into the form (x + a)² or (x – a)². To find this value, we take half of the coefficient of our x term (which is 8), square it ((8/2)² = 16), and then add and subtract it inside the parentheses: F(x) = -(x² + 8x + 16 – 16) + 4. We're essentially adding zero here, so we're not changing the value of the function.
  4. Rewrite as a perfect square: Now we can rewrite the expression inside the parentheses as a perfect square: F(x) = -((x + 4)² – 16) + 4. See how x² + 8x + 16 neatly becomes (x + 4)²?
  5. Distribute and simplify: Next, distribute the -1 to both terms inside the parentheses: F(x) = -(x + 4)² + 16 + 4. Then, combine the constants: F(x) = -(x + 4)² + 20.

Ta-da! We've completed the square! Our function is now in the form F(x) = a(x – h)² + k. Now we can easily identify the vertex.

Identifying the Vertex

Now that we have F(x) = -(x + 4)² + 20, we can directly read off the vertex. Remember, the vertex is (h, k) in the form F(x) = a(x – h)² + k. Notice that the 'h' value has a negative sign in the formula, so we need to be careful. In our case, we have (x + 4) inside the parentheses, which is the same as (x – (-4)). So, h = -4, and k = 20. Therefore, the vertex of our parabola is (-4, 20).

This is a huge result! The vertex is the turning point of the parabola. Since we know our parabola opens downwards (because the coefficient of x² is negative), the vertex represents the maximum point on the graph. The y-coordinate of the vertex, 20, is the maximum y-value that our function can achieve.

Determining the Range

Alright, we've done all the hard work, and now we can easily determine the range. We know the parabola opens downwards, and the vertex is (-4, 20), which means the maximum y-value is 20. Since the parabola extends downwards indefinitely, the range includes all y-values less than or equal to 20.

We can express this range in a couple of ways:

  • Interval notation: (-∞, 20]
  • Set notation: {y | y ≤ 20}

Both notations mean the same thing: the range includes all real numbers less than or equal to 20. The square bracket on the 20 indicates that 20 is included in the range, while the parenthesis on the -∞ indicates that negative infinity is not included (since infinity isn't a number).

Putting It All Together

Let's recap the steps we took to find the range of F(x) = 4 – 8x – x²:

  1. Understand functions and range: We defined what a function and its range are, and how they relate to quadratic functions.
  2. Identify the type of quadratic function: We determined that our parabola opens downwards because the coefficient of x² is negative.
  3. Complete the square: We used the completing the square method to rewrite the function in vertex form: F(x) = -(x + 4)² + 20.
  4. Identify the vertex: We identified the vertex as (-4, 20) from the vertex form.
  5. Determine the range: We used the vertex and the direction of the parabola to determine the range as (-∞, 20] or {y | y ≤ 20}.

Practice Makes Perfect

Finding the range of quadratic functions might seem tricky at first, but with practice, you'll become a pro! The key is to understand the connection between the parabola's shape, the vertex, and the range. Completing the square is a powerful tool, so make sure you're comfortable with that technique.

Try working through a few more examples on your own. Change the coefficients and constants in the quadratic function and see if you can find the range. The more you practice, the more confident you'll become. And remember, if you get stuck, just break it down step by step, and you'll get there!

So, there you have it! Figuring out the range of F(x) = 4 – 8x – x² isn't so scary after all. Keep practicing, and you'll be a quadratic function whiz in no time!