Understanding The Range Of F(x) = -(x+3)^2+7

Hey guys! Today we're diving into a super common question in math, especially when you're getting your head around functions: what is the range of the function $f(x)=-(x+3)^2+7$? This might seem a bit intimidating with the squared term and the negative sign, but trust me, once you break it down, it's pretty straightforward. We're going to figure out which output values, or y-values, this function can actually produce. So, grab your notebooks, maybe a snack, and let's get this solved! Understanding the range is crucial because it tells us the set of all possible output values for a given function. For $f(x)=-(x+3)^2+7$, we're essentially looking for all the possible y-values that the graph of this function can reach. The standard form of a quadratic function is $ax^2+bx+c$, but this one is presented in vertex form, which is $a(x-h)^2+k$. This form is super helpful because it directly tells us the vertex of the parabola, which is at $(h, k)$. In our case, the function is $f(x)=-(x+3)^2+7$. Comparing this to $a(x-h)^2+k$, we can see that $a = -1$, $h = -3$, and $k = 7$. So, the vertex of our parabola is at $(-3, 7)$. Now, the value of '$a$' tells us the direction the parabola opens. Since $a = -1$ (which is negative), our parabola opens downwards. This is a really key piece of information, guys. When a parabola opens downwards, its highest point is its vertex. After that point, the graph just keeps going down infinitely. This means the maximum value the function can ever reach is the y-coordinate of the vertex. In our function, the y-coordinate of the vertex is 7. Therefore, the function can produce any y-value that is 7 or less than 7. It can never go above 7 because the vertex is the absolute highest point. This is how we start to define the range. We know it's all real numbers less than or equal to 7. The other part of the range is determined by the fact that it opens downwards, meaning it extends infinitely in the negative y direction. So, the range is not just any number less than or equal to 7, but all real numbers less than or equal to 7. This covers the entire spectrum of possible outputs for this specific quadratic function. It's all about understanding how the components of the vertex form influence the graph's position and orientation.

Deconstructing the Function: The Vertex Form is Your Best Friend

Alright, let's get a bit more granular with this function, $f(x)=-(x+3)^2+7$. We've already identified that it's in vertex form, $a(x-h)^2+k$, which is seriously a lifesaver for understanding parabolas. Remember, the vertex $(h, k)$ is a critical point, and in our case, it's $(-3, 7)$. Now, let's talk about the coefficient '$a$', which is $-1$ here. The sign of '$a$' is huge because it dictates whether the parabola opens upwards or downwards. Since '$a$' is negative, our parabola is like a frown – it opens downwards. This downward opening is what limits the maximum output of the function. Imagine the graph: it goes up and up, hits the vertex at $(-3, 7)$, and then starts coming down, heading towards negative infinity. It will never go higher than the y-value of the vertex, which is 7. This is why the range is constrained from above. The term $(x+3)^2$ is always going to be greater than or equal to zero, because any real number squared is non-negative. The smallest value $(x+3)^2$ can take is 0, which happens when $x = -3$. When $(x+3)^2$ is 0, our function becomes $f(-3) = -(0) + 7 = 7$. This confirms that the maximum value is indeed 7, occurring at $x=-3$. Now, what happens when $(x+3)^2$ gets bigger? Well, since we have a negative sign in front of it, $-(x+3)^2$ becomes a negative number, and the larger $(x+3)^2$ gets, the more negative $-(x+3)^2$ becomes. For example, if $x=0$, then $(x+3)^2 = 3^2 = 9$, and $f(0) = -(9) + 7 = -2$. If $x=7$, then $(x+3)^2 = 10^2 = 100$, and $f(7) = -(100) + 7 = -93$. As you can see, as $|x+3|$ increases (meaning $x$ gets further away from -3 in either the positive or negative direction), the value of $-(x+3)^2$ becomes a larger and larger negative number. Adding 7 to this increasingly negative number means the function's output, $f(x)$, will decrease indefinitely. It will approach negative infinity. So, the function can take on any value from 7 downwards. This confirms that the range includes 7 and all real numbers smaller than 7. Therefore, the range is all real numbers less than or equal to 7. It's like saying the function's y-values can be $7, 6.9, 0, -100, -1000000$, and so on, but never $7.1$ or $8$. It's pretty neat how the vertex form lays all this out for us!

Connecting to the Options: Finding the Correct Answer

So, we've done the heavy lifting, guys. We've analyzed the function $f(x)=-(x+3)^2+7$, identified its vertex form, determined the vertex coordinates, and figured out the direction it opens. Let's recap what we found: the vertex is at $(-3, 7)$, and because the coefficient '$a$' is $-1$ (negative), the parabola opens downwards. This means the highest point the function reaches is the y-coordinate of the vertex, which is 7. Consequently, the function can output any y-value that is less than or equal to 7. It cannot output any value greater than 7. Now, let's look at the options provided to see which one matches our findings:

• A. all real numbers less than or equal to 7
• B. all real numbers greater than or equal to 7
• C. all real numbers less than or equal to -3
• D. all real numbers greater than or equal to -3

Based on our analysis, the range of the function is the set of all possible y-values, which we determined to be all real numbers less than or equal to 7. This perfectly matches Option A. Option B is incorrect because the parabola opens downwards, meaning it has a maximum value, not a minimum value greater than or equal to 7. The values are less than or equal to 7, not greater than. Option C and D are talking about the domain, which is the set of all possible x-values. For any quadratic function, the domain is typically all real numbers unless stated otherwise. The value -3 is the x-coordinate of the vertex, which influences where the maximum occurs, but it's not the range itself. The range is about the output (y-values), not the input (x-values). So, to be crystal clear, the range of $f(x)=-(x+3)^2+7$ is indeed all real numbers less than or equal to 7. This conclusion stems directly from understanding the vertex form and the impact of the leading coefficient on the parabola's orientation. Keep practicing these types of problems, and you'll become a pro at spotting the range (and domain!) in no time. It's all about building that mathematical intuition, and breaking down functions into their core components is the best way to do it, guys!

Why Other Options Are Incorrect: A Deeper Dive

Let's really hammer home why the other options just don't cut it for the range of $f(x)=-(x+3)^2+7$. We've established that the range represents the set of all possible output values, or y-values, that a function can produce. For our function, the vertex is at $(-3, 7)$, and it opens downwards because the leading coefficient ($a=-1$) is negative. This means 7 is the absolute highest y-value the function will ever reach. Any y-value greater than 7 is simply not possible for this function. This immediately disqualifies Option B: all real numbers greater than or equal to 7. If the parabola opened upwards (meaning $a$ was positive), then 7 would be the minimum value, and the range would be all real numbers greater than or equal to 7. But that's not the case here.

Now, let's consider Option C: all real numbers less than or equal to -3 and Option D: all real numbers greater than or equal to -3. These options are talking about the number -3. What does -3 represent in our function? It's the x-coordinate of the vertex. The x-coordinate of the vertex tells us the input value where the maximum (or minimum, if the parabola opened upwards) occurs. The set of all possible input values (x-values) is called the domain. For most quadratic functions like this one, the domain is all real numbers, because you can plug any real number into $x$ and get a valid output. Options C and D describe intervals of x-values. For example, Option D would mean that $x$ can be any number like $-3, -2, 0, 5, 100$, etc. Option C would mean $x$ can be $-3, -4, -10, -100$, etc. While the function does produce outputs for these x-values, these options are describing the domain, not the range. The range is exclusively about the y-values. The fact that the maximum y-value occurs at $x=-3$ is important, but it doesn't mean the range itself is related to the number -3 in this way. The range is determined by the y-coordinate of the vertex and the direction the parabola opens. Since the y-coordinate of the vertex is 7 and it opens downwards, the range is all y-values less than or equal to 7. So, you see, it's crucial to distinguish between domain and range. The domain is about what numbers you can put into the function, and the range is about what numbers you can get out of the function. By carefully analyzing the vertex and the direction of opening, we can confidently say that Option A is the only correct description of the range for $f(x)=-(x+3)^2+7$. It's a common mistake to confuse domain and range or to misinterpret the role of the vertex coordinates, but with a little practice and careful thought, you guys will master it!