Mastering $f(x)=(x-5)^2+1$: Your Easy Graphing Guide

Ever wondered how those graceful, curvy lines in math textbooks come to life? We're talking about parabolas, the visual representation of quadratic functions, which pop up everywhere from the trajectory of a thrown ball to the design of satellite dishes. If you've ever felt a bit daunted by the idea of graphing functions or completing a function table, you're in the right place! Today, we're going to dive into a specific, yet very common, type of quadratic function: $f(x)=(x-5)^2+1$. This article will serve as your friendly, comprehensive guide to not only understanding this particular function but also meticulously completing its function table for a given domain and then accurately plotting the points on a graph. By the end of our journey together, you'll feel much more confident in tackling similar problems, armed with the knowledge and practical steps needed to turn abstract equations into beautiful, tangible graphs. We'll break down the concepts into bite-sized pieces, ensuring that each step is clear, easy to follow, and incredibly valuable. This isn't just about solving a math problem; it's about building a foundational understanding of how functions behave and how their algebraic expressions translate into visual patterns, which is a truly empowering skill in mathematics and beyond. So, grab your pencil, some graph paper, and let's embark on this exciting mathematical adventure together, transforming an equation into a visual masterpiece that reveals its inherent characteristics and behavior. This exploration will show you just how accessible and even fun understanding quadratic equations and their graphical representations can be, making complex topics seem simple and engaging.

Unpacking the Quadratic Function: $f(x)=(x-5)^2+1$

Before we jump into calculations, let's take a moment to really understand what we're looking at with $f(x)=(x-5)^2+1$. This isn't just a random jumble of numbers and symbols; it's a quadratic function presented in what mathematicians call vertex form. The vertex form of a quadratic equation is generally written as $f(x) = a(x-h)^2 + k$, and it's super handy because it immediately tells us a lot about the parabola's shape and position. In our specific function, $f(x)=(x-5)^2+1$, we can directly identify some key pieces of information. The 'a' value here is implicitly 1 (since there's no number multiplying the squared term), which tells us the parabola opens upwards. If 'a' were negative, it would open downwards. More importantly, the 'h' and 'k' values give us the exact coordinates of the parabola's vertex. The vertex is that crucial turning point of the parabola – either its lowest point (if it opens up) or its highest point (if it opens down). For $f(x)=(x-5)^2+1$, our 'h' is 5 (because of the $(x-5)$ part, remember it's $(x-h)$ so we take the opposite sign), and our 'k' is 1. Voila! We know the vertex of this parabola is at the point (5, 1). This is a powerful piece of information because it gives us a starting point and a central reference for our graph. The term $(x-5)^2$ is what makes this a quadratic function; the exponent of 2 means that as x changes, f(x) will change in a non-linear, squared fashion, producing that characteristic parabolic curve. The subtraction of 5 inside the parenthesis tells us that the standard $y=x^2$ parabola has been shifted 5 units to the right (again, opposite of what you might initially think for subtraction). Then, the addition of 1 outside the parenthesis shifts the entire parabola 1 unit upwards. So, this function is essentially a basic $y=x^2$ parabola that has been gracefully moved to a new home on the coordinate plane. Understanding these transformations is key to visualizing the graph even before plotting a single point, providing valuable insight into the overall structure and behavior of our function. It's truly amazing how much information is encoded in such a seemingly simple equation, making its graphical representation a predictable and logical outcome of its algebraic form, reinforcing the interconnectedness of algebra and geometry.

Step-by-Step: Completing Your Function Table

Now that we're friends with our function, $f(x)=(x-5)^2+1$, it's time to get down to the practical task of completing the function table. This table is our roadmap to plotting the points on the graph. We've been given a specific domain, which means we only need to calculate f(x) for a handful of x-values: 2, 3, 4, 5, and 6. This is a fantastic way to start understanding how the function behaves within a particular range. For each x-value, we'll simply substitute it into our function and carefully calculate the corresponding f(x) value. Let's walk through it together, one step at a time, ensuring precision in every calculation.

First, let's consider $x = 2$: $f(2) = (2-5)^2 + 1$ $f(2) = (-3)^2 + 1$ $f(2) = 9 + 1$ $f(2) = 10$ So, our first ordered pair is (2, 10).

Next, for $x = 3$: $f(3) = (3-5)^2 + 1$ $f(3) = (-2)^2 + 1$ $f(3) = 4 + 1$ $f(3) = 5$ This gives us the ordered pair (3, 5).

Moving on to $x = 4$: $f(4) = (4-5)^2 + 1$ $f(4) = (-1)^2 + 1$ $f(4) = 1 + 1$ $f(4) = 2$ Our third pair is (4, 2).

Now, for $x = 5$: This is a special one, remember our vertex? $f(5) = (5-5)^2 + 1$ $f(5) = (0)^2 + 1$ $f(5) = 0 + 1$ $f(5) = 1$ Indeed, this confirms our vertex at (5, 1)! This is a crucial point for our graph.

Finally, for $x = 6$: $f(6) = (6-5)^2 + 1$ $f(6) = (1)^2 + 1$ $f(6) = 1 + 1$ $f(6) = 2$ Our last ordered pair is (6, 2).

See how straightforward that was? The key is to follow the order of operations (PEMDAS/BODMAS) meticulously. Parentheses first, then exponents, then addition. Now, let's consolidate our findings into a neat table, which will make our plotting points step incredibly easy:

Notice the beautiful symmetry around our vertex at $x=5$. The f(x) values for $x=4$ and $x=6$ are both 2, and for $x=3$ and $x=7$ (if we had included it), they would be 5. This symmetry is a hallmark of parabolas and a great way to double-check your calculations, reinforcing your understanding of quadratic function behavior. This table now gives us a clear set of ordered pairs that we can confidently transfer onto our graph paper, bringing the function to life visually.

Plotting the Points: Bringing Your Parabola to Life

With our meticulously completed function table in hand, we're now ready for the exciting part: plotting the points and watching our parabola take shape! This step transforms abstract numbers into a tangible, visual representation, which is often the most intuitive way to understand a function's behavior. To do this, you'll need a piece of graph paper and a pencil. Remember, each row in our table gives us an ordered pair in the format (x, f(x)), which corresponds directly to (x, y) coordinates on a coordinate plane.

Let's list our ordered pairs again, just to keep them fresh in our minds:

• (2, 10)
• (3, 5)
• (4, 2)
• (5, 1)
• (6, 2)

To plot these points, first, draw your x-axis (horizontal) and y-axis (vertical) on your graph paper, making sure they intersect at the origin (0,0). Label your axes clearly and choose an appropriate scale. Since our x-values range from 2 to 6 and our y-values range from 1 to 10, a scale where each grid line represents one unit will work perfectly for both axes. Now, let's locate each point:

1. Point (2, 10): Start at the origin, move 2 units to the right along the x-axis, then move 10 units up parallel to the y-axis. Place a clear dot there.
2. Point (3, 5): From the origin, move 3 units right, then 5 units up. Mark it.
3. Point (4, 2): Go 4 units right, then 2 units up. Plot this point.
4. Point (5, 1): Move 5 units right, then 1 unit up. This is our vertex! It's the lowest point on this parabola. Mark it prominently.
5. Point (6, 2): Finally, go 6 units right, then 2 units up. Place your last dot.

Once all five points are plotted, you'll start to see a distinct curve emerging. Do not connect the dots with straight lines! A parabola is a smooth curve. Take your pencil and, starting from one end, gently draw a smooth, continuous curve that passes through all the points. It should look like a U-shape opening upwards. The point (5, 1) will be the very bottom of this