Find The Function From Its X-Intercepts

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of mathematics, specifically tackling a super common problem: finding the right function when you're given its x-intercepts. You know, those points where the graph of a function crosses the x-axis? They're super important for understanding a function's behavior, and knowing how to work with them can really boost your math game. So, let's get this party started and figure out which function fits the bill based on the intercepts provided.

Understanding X-Intercepts and Functions

Alright, let's get down to brass tacks, fam. What exactly are x-intercepts, and why should we care? In plain English, an x-intercept is a point on a graph where the y-value is zero. Think of it as the spot where the function 'hits' the x-axis. Mathematically, if a function $g(x)$ has an x-intercept at $(a, 0)$, it means that when you plug in $x=a$ into the function, the output $g(a)$ is equal to 0. This is a huge clue about the function's structure. For polynomial functions, especially quadratics, these x-intercepts are directly related to the factors of the polynomial. Remember the Factor Theorem? It tells us that if $(a, 0)$ is an x-intercept of a polynomial $g(x)$, then $(x-a)$ must be a factor of $g(x)$. This theorem is our secret weapon for solving problems like this. So, when we see x-intercepts at (rac{1}{2}, 0) and $(6, 0)$, we immediately know that (x - rac{1}{2}) and $(x - 6)$ must be factors of our function $g(x)$. Now, a function can have a leading coefficient other than 1, so our function might look something like g(x) = k(x - rac{1}{2})(x - 6), where 'k' is some non-zero constant. The value of 'k' affects the stretch or compression of the graph, and whether it opens upwards or downwards, but it doesn't change the x-intercepts themselves. Our mission, should we choose to accept it, is to find the option that includes these factors, possibly multiplied by a constant and maybe with some clever algebraic manipulation to get rid of fractions.

Analyzing the Given X-Intercepts

So, we're given that our function $g(x)$ has x-intercepts at (rac{1}{2}, 0) and $(6, 0)$. Using the power of the Factor Theorem we just talked about, this means that (x - rac{1}{2}) and $(x - 6)$ are factors of $g(x)$. So, we can write $g(x)$ in the form g(x) = k(x - rac{1}{2})(x - 6) for some constant $k$. Now, looking at the options, they are all presented in a factored form, but some might look a little different. We need to be sharp and see which one matches our understanding. Option A has factors $(x+1)$ and $(x+6)$, which would give x-intercepts at $(-1, 0)$ and $(-6, 0)$. That's a definite no-go, guys. Option C has factors $(x-2)$ and $(x-6)$. This gives an x-intercept at $(6, 0)$, which is good, but the other intercept is at $(2, 0)$, not (rac{1}{2}, 0). So, Option C is also out. Option D has factors $(x+6)$ and $(x+2)$, giving x-intercepts at $(-6, 0)$ and $(-2, 0)$. Again, not what we're looking for. Now, let's look closely at Option B: $g(x) = (x - 6)(2x - 1)$. We already know that $(x - 6)$ is a factor, which correctly gives us the x-intercept at $(6, 0)$. What about the factor $(2x - 1)$? If we set this factor equal to zero to find its corresponding x-intercept, we get $2x - 1 = 0$. Solving for $x$, we add 1 to both sides to get $2x = 1$, and then divide by 2 to get x = rac{1}{2}. So, the x-intercept from this factor is (rac{1}{2}, 0)! Bingo! Option B has the exact x-intercepts we were given. It's like it was made for this problem. But wait, is it in the form k(x - rac{1}{2})(x - 6)? Let's check. If we take Option B, $g(x) = (x - 6)(2x - 1)$, and distribute the terms, we get $g(x) = 2x^2 - x - 12x + 6 = 2x^2 - 13x + 6$. Now, let's consider our general form g(x) = k(x - rac{1}{2})(x - 6). If we expand this, we get g(x) = k(x^2 - 6x - rac{1}{2}x + 3) = k(x^2 - rac{13}{2}x + 3). To make this match $2x^2 - 13x + 6$, we can see that if $k=2$, then g(x) = 2(x^2 - rac{13}{2}x + 3) = 2x^2 - 13x + 6. So, Option B is indeed equivalent to 2(x - rac{1}{2})(x - 6), which clearly shows the factors corresponding to the given x-intercepts. The key here is recognizing that the factor $(2x - 1)$ is just a scaled version of (x - rac{1}{2}). Specifically, 2x - 1 = 2(x - rac{1}{2}). So, the function $g(x) = (x - 6)(2x - 1)$ can be rewritten as g(x) = (x - 6) imes 2(x - rac{1}{2}), which is g(x) = 2(x - rac{1}{2})(x - 6). This confirms that Option B is the correct function.

Evaluating Each Option Systematically

Let's break down each option piece by piece to make sure we're not missing anything, guys. This systematic approach is crucial for accuracy in math. We're given that the x-intercepts are (rac{1}{2}, 0) and $(6, 0)$. This means that g(rac{1}{2}) = 0 and $g(6) = 0$. We'll plug these values into each option and see which one holds true for both conditions.

Option A: $g(x) = 2(x+1)(x+6)$

If $g(x) = 2(x+1)(x+6)$, let's test our intercepts:

For x = rac{1}{2}: g(rac{1}{2}) = 2(rac{1}{2} + 1)(rac{1}{2} + 6) = 2(rac{3}{2})(rac{13}{2}) = rac{3 imes 13}{2} = rac{39}{2}. This is definitely not 0. So, Option A is incorrect.

For $x = 6$: $g(6) = 2(6+1)(6+6) = 2(7)(12) = 168$. This is also not 0.

As we suspected, Option A does not have the correct x-intercepts.

Option B: $g(x) = (x-6)(2x-1)$

Now for the contender, Option B. Let's test our intercepts:

For x = rac{1}{2}: g(rac{1}{2}) = (rac{1}{2} - 6)(2(rac{1}{2}) - 1) = (-rac{11}{2})(1 - 1) = (-rac{11}{2})(0) = 0. Success! This point works.

For $x = 6$: $g(6) = (6 - 6)(2(6) - 1) = (0)(12 - 1) = (0)(11) = 0$. Success again! This point also works.

Since both x-intercepts satisfy the function in Option B, this is our likely answer. But let's be thorough and check the others just to be absolutely sure.

Option C: $g(x) = 2(x-2)(x-6)$

Let's test the intercepts with Option C:

For x = rac{1}{2}: g(rac{1}{2}) = 2(rac{1}{2} - 2)(rac{1}{2} - 6) = 2(-rac{3}{2})(-rac{11}{2}) = rac{2 imes (-3) imes (-11)}{4} = rac{66}{4} = rac{33}{2}. This is not 0.

For $x = 6$: $g(6) = 2(6 - 6)(6 - 6) = 2(0)(0) = 0$. The second intercept works, but the first one doesn't.

Therefore, Option C is incorrect.

Option D: $g(x) = (x+6)(x+2)$

Finally, let's check Option D:

For x = rac{1}{2}: g(rac{1}{2}) = (rac{1}{2} + 6)(rac{1}{2} + 2) = (rac{13}{2})(rac{5}{2}) = rac{65}{4}. This is not 0.

For $x = 6$: $g(6) = (6 + 6)(6 + 2) = (12)(8) = 96$. This is also not 0.

Option D clearly does not have the required x-intercepts.

The Factor Theorem in Action: Why Option B Works

As we've seen through systematic testing, Option B is the only one that satisfies both x-intercept conditions. But let's circle back to the Factor Theorem to really understand why it works. The theorem states that if $(a, 0)$ is an x-intercept, then $(x-a)$ is a factor. So, for x-intercepts (rac{1}{2}, 0) and $(6, 0)$, we expect factors (x - rac{1}{2}) and $(x - 6)$.

Option B is $g(x) = (x - 6)(2x - 1)$. We already confirmed that $(x-6)$ is a factor. Now, consider the factor $(2x - 1)$. If we set $2x - 1 = 0$, we solve for $x$ and get x = rac{1}{2}. This means that $(2x - 1)$ is indeed the factor corresponding to the x-intercept (rac{1}{2}, 0).

It's important to note that the factor doesn't have to be in the form $(x-a)$. It can be $k(x-a)$ for any non-zero constant $k$. In Option B, the factor $(2x - 1)$ is equivalent to 2(x - rac{1}{2}). So, we can rewrite Option B as:

g(x) = (x - 6) imes [2(x - rac{1}{2})]

g(x) = 2(x - rac{1}{2})(x - 6)

This form explicitly shows the factors corresponding to our x-intercepts, with a leading coefficient of $k=2$. This perfectly matches our expectation based on the Factor Theorem. The other options had factors that led to different x-intercepts, making them incorrect. This reinforces the power of understanding fundamental mathematical theorems like the Factor Theorem.

Conclusion: Mastering Function Intercepts

So, there you have it, math whizzes! When faced with a problem asking you to identify a function from its x-intercepts, remember the golden rule: the x-intercept $(a, 0)$ implies that $(x-a)$ is a factor of the function. Keep an eye out for variations, like $(kx - m)$ which is just a scaled version of (x - rac{m}{k}). By applying the Factor Theorem and systematically checking each option, you can confidently pinpoint the correct function. It's all about understanding the relationship between roots (where the function equals zero) and factors. Keep practicing these concepts, and you'll be a function-finding pro in no time. Stay curious, keep learning, and we'll catch you in the next article!