Find The Function: Passing Through Key Points

Hey guys! Ever been faced with a math problem that feels like a puzzle? You've got a few pieces of information, like a graph passing through specific points, and you need to figure out the whole picture – the function itself! Today, we're diving deep into a classic math challenge: identifying a function given a set of points it passes through. This is super common in algebra and pre-calculus, and understanding how to tackle it will make you feel like a math whiz. We'll break down a specific problem that involves points (0,2), (1,0), and (-1,0) and explore how to test different function options to find the right one. Get ready to flex those problem-solving muscles!

Understanding the Problem: Points on a Graph

So, what does it mean for a graph of a function $f(x)$ to pass through certain points? It's actually pretty straightforward, guys. When we say a graph passes through a point $(x, y)$, it simply means that if you plug the $x$-value into the function, you get the corresponding $y$-value as the output. In mathematical terms, $f(x) = y$. Our problem gives us three such points: $(0,2)$, $(1,0)$, and $(-1,0)$. This means three conditions must be met by the correct function $f(x)$. Let's list them out:

1. $f(0) = 2$
2. $f(1) = 0$
3. $f(-1) = 0$

We're then presented with four possible functions (A, B, C, and D) and we need to figure out which one satisfies all three of these conditions. This is where the detective work begins! We'll systematically test each function against each point. It’s like trying on different outfits to see which one fits perfectly. Some functions might fit one or two points, but only the correct one will fit all three. This method is called substitution or evaluation, and it's a fundamental technique in mathematics.

Testing the Options: A Step-by-Step Approach

Let's tackle each option one by one. We'll be substituting the $x$-values from our points into each function and checking if the resulting $y$-value matches the one given in the point. If even one point doesn't match for a function, we can rule that function out. It's a process of elimination, and we're aiming to eliminate the incorrect ones until only the correct function remains.

Option A: $f(x) = -2x^2 + 2$

• Test point (0,2): $f(0) = -2(0)^2 + 2 = -2(0) + 2 = 0 + 2 = 2$. Success! This point matches.
• Test point (1,0): $f(1) = -2(1)^2 + 2 = -2(1) + 2 = -2 + 2 = 0$. Success! This point also matches.
• Test point (-1,0): $f(-1) = -2(-1)^2 + 2 = -2(1) + 2 = -2 + 2 = 0$. Success! This point matches too!

Wow, Option A seems to work for all three points! Before we declare victory, let's quickly check the others just to be absolutely sure and to see how they fail.

Option B: $f(x) = -2x + 2$

• Test point (0,2): $f(0) = -2(0) + 2 = 0 + 2 = 2$. Success!
• Test point (1,0): $f(1) = -2(1) + 2 = -2 + 2 = 0$. Success!
• Test point (-1,0): $f(-1) = -2(-1) + 2 = 2 + 2 = 4$. Failure! This point should be 0, but we got 4.

So, Option B is out. It passes through (0,2) and (1,0), but not (-1,0).

Option C: $f(x) = 2x + 2$

• Test point (0,2): $f(0) = 2(0) + 2 = 0 + 2 = 2$. Success!
• Test point (1,0): $f(1) = 2(1) + 2 = 2 + 2 = 4$. Failure! This point should be 0, but we got 4.

Option C is also out. It only passes through (0,2).

Option D: $f(x) = -2\sqrt{x} + 2$

• Test point (0,2): $f(0) = -2\\sqrt{0} + 2 = -2(0) + 2 = 0 + 2 = 2$. Success!
• Test point (1,0): $f(1) = -2\\sqrt{1} + 2 = -2(1) + 2 = -2 + 2 = 0$. Success!
• Test point (-1,0): Here's a major issue, guys. The square root of a negative number is not a real number. So, $f(-1)$ is undefined in the real number system. This means the function cannot pass through the point (-1,0) if we're working with real-valued functions, which is standard unless specified otherwise.

Therefore, Option D is definitely out due to domain restrictions for real numbers.

The Winner Is...

After testing all the options, we found that Option A: $f(x) = -2x^2 + 2$ is the only function that satisfies all three given points: $(0,2)$, $(1,0)$, and $(-1,0)$. This quadratic function perfectly models the given data points. It's awesome when a plan comes together, right?

Why This Matters: Connecting Functions and Graphs

Understanding how to verify if points lie on a function's graph is a fundamental skill in mathematics. It's not just about solving problems; it's about building an intuition for how functions behave. For instance, in Option A, $f(x) = -2x^2 + 2$, we see a quadratic function. The $-2x^2$ term tells us the parabola opens downwards, and the $+2$ shifts it upwards by 2 units. This matches the behavior suggested by the points: starting at y=2 when x=0, and then decreasing to y=0 at x=1 and x=-1. The symmetry around the y-axis (since $f(x) = f(-x)$ for this function) is also evident from the points $(1,0)$ and $(-1,0)$ having the same y-value.

In contrast, Option B and C, $f(x) = -2x + 2$ and $f(x) = 2x + 2$, are linear functions. Linear functions produce straight lines. A single straight line can only pass through at most two distinct points unless those points are collinear. Since we have three points, and they aren't collinear in a way that fits these specific linear functions, they fail. Option D, $f(x) = -2\\sqrt{x} + 2$, involves a square root. The standard square root function, $\\sqrt{x}$, is only defined for non-negative $x$ values. This domain restriction immediately disqualifies it from passing through $(-1,0)$ in the real number system. If we were dealing with complex numbers, the scenario might change, but in typical high school or introductory college math, we stick to real numbers.

This exercise really highlights the importance of function properties and their graphical representations. Each type of function – linear, quadratic, exponential, logarithmic, trigonometric, radical – has unique characteristics that dictate the shape of its graph and the points it can pass through. Being able to test points is like having a key to unlock the identity of a function. It’s a skill that will serve you well as you encounter more complex mathematical concepts. Keep practicing, and you’ll get faster and more confident in identifying functions and understanding their behavior. Math on!