Finding Functions With Specific Zeros

Hey guys, let's dive into a super common math problem: figuring out which function has specific zeros. You know, those values of x that make the function equal to zero. Today, we're tackling a question about finding a function with zeros at $x=10$ and $x=2$. This might sound a bit tricky, but trust me, once you get the hang of it, it's a piece of cake! We'll break down how to approach this, explore the given options, and figure out the right answer together. So, grab your notebooks, and let's get this math party started!

Understanding Zeros of a Function

Alright, let's talk zeros. What are they, really? In simple terms, the zeros of a function, also known as roots, are the input values (x-values) that make the function's output (f(x) or y-value) equal to zero. Think of it like this: if you graph a function, the zeros are the points where the graph crosses or touches the x-axis. These are super important because they tell us a lot about the behavior of the function. For a polynomial function, like the quadratic functions we're dealing with here, if you know the zeros, say at $x=a$ and $x=b$, you can actually write the function in a factored form: $f(x) = k(x-a)(x-b)$, where k is just some constant multiplier. This is a powerful concept, guys, because it means we can work backward from the zeros to find the function, or vice versa. For our specific problem, we're given that the zeros are $x=10$ and $x=2$. Using our factored form idea, the function should look something like $f(x) = k(x-10)(x-2)$. Now, the different answer choices are quadratic functions, which are polynomials of degree two. They have the general form $f(x) = ax^2 + bx + c$. The trick is to see which of the given options, when expanded and simplified, matches the form derived from the zeros. We'll be using the Factor Theorem here, which is a fundamental concept in algebra. It states that a polynomial $P(x)$ has a zero at $x=c$ if and only if $(x-c)$ is a factor of $P(x)$. So, if $x=10$ is a zero, then $(x-10)$ must be a factor. If $x=2$ is a zero, then $(x-2)$ must be a factor. Therefore, our function must have both $(x-10)$ and $(x-2)$ as factors. This means the function should be proportional to the product of these two factors: $(x-10)(x-2)$. Let's expand this product: $(x-10)(x-2) = x^2 - 2x - 10x + 20 = x^2 - 12x + 20$. So, any function that has zeros at $x=10$ and $x=2$ must be of the form $f(x) = k(x^2 - 12x + 20)$ for some non-zero constant k. Our job now is to check the given options to see which one fits this form. Keep in mind that the constant k could be 1, or it could be something else, like 5 in some of the options. We're basically looking for a function that, when factored, yields $(x-10)$ and $(x-2)$.

Analyzing the Options: A Step-by-Step Approach

Okay team, let's break down each of the answer choices provided to see which one fits our criteria. Remember, we're looking for a function where $f(10) = 0$ and $f(2) = 0$. We can either plug in the values of x directly, or we can try to factor each quadratic equation to see if it yields the factors $(x-10)$ and $(x-2)$. Let's try the factoring method first, as it's generally more insightful. Remember our goal is to find a function of the form $f(x) = k(x^2 - 12x + 20)$.

Option A: $f(x) = x^2 - 12x + 20$

This looks promising right off the bat! Let's see if we can factor this quadratic. We need two numbers that multiply to 20 and add up to -12. Think about the pairs of factors for 20: (1, 20), (2, 10), (4, 5). To get a sum of -12, we need both numbers to be negative. So, let's consider (-1, -20), (-2, -10), (-4, -5). Bingo! -2 and -10 multiply to 20 and add up to -12. So, we can factor this as $f(x) = (x - 2)(x - 10)$. This means the zeros are indeed $x=2$ and $x=10$. This option looks like our winner, guys! The constant k here is 1.

Option B: $f(x) = x^2 - 20x + 12$

Let's try to factor this one. We need two numbers that multiply to 12 and add up to -20. Pairs of factors for 12 are (1, 12), (2, 6), (3, 4). To get a negative sum, we'd need both to be negative: (-1, -12), (-2, -6), (-3, -4). Let's check the sums: -1 + (-12) = -13, -2 + (-6) = -8, -3 + (-4) = -7. None of these add up to -20. This quadratic doesn't seem to factor nicely into integers that would give us the zeros we need. We could also check the discriminant ($\Delta = b^2 - 4ac$) to see if it has real roots, but even if it does, they are unlikely to be 10 and 2. For this function, $a=1, b=-20, c=12$. The discriminant is $(-20)^2 - 4(1)(12) = 400 - 48 = 352$. Since 352 is not a perfect square, the roots are irrational, so they definitely won't be 10 and 2. So, Option B is out.

Option C: $f(x) = 5x^2 + 40x + 60$

First, let's see if we can factor out a common factor. All the coefficients (5, 40, 60) are divisible by 5. So, we can rewrite this as $f(x) = 5(x^2 + 8x + 12)$. Now, let's focus on the quadratic inside the parentheses: $x^2 + 8x + 12$. We need two numbers that multiply to 12 and add up to 8. Let's look at the factors of 12: (1, 12), (2, 6), (3, 4). The sums are 1+12=13, 2+6=8, 3+4=7. Perfect! The numbers 2 and 6 work. So, $x^2 + 8x + 12$ factors into $(x+2)(x+6)$. This means $f(x) = 5(x+2)(x+6)$. The zeros of this function are where $(x+2)=0$ or $(x+6)=0$, which gives us $x=-2$ and $x=-6$. These are not the zeros we're looking for ($x=10$ and $x=2$). So, Option C is incorrect.

Option D: $f(x) = 5x^2 + 60x + 100$

Again, let's factor out the common factor, which is 5. $f(x) = 5(x^2 + 12x + 20)$. Now, let's factor the quadratic $x^2 + 12x + 20$. We need two numbers that multiply to 20 and add up to 12. Factors of 20 are (1, 20), (2, 10), (4, 5). Let's check the sums: 1+20=21, 2+10=12, 4+5=9. Awesome! The numbers 2 and 10 work. So, $x^2 + 12x + 20$ factors into $(x+2)(x+10)$. This means $f(x) = 5(x+2)(x+10)$. The zeros of this function are where $(x+2)=0$ or $(x+10)=0$, which gives us $x=-2$ and $x=-10$. These are also not the zeros we're looking for. So, Option D is incorrect.

The Correct Answer and Why

After carefully analyzing each option, it's clear that Option A: $f(x) = x^2 - 12x + 20$ is the function that has zeros at $x=10$ and $x=2$. We found this by factoring the quadratic. The factors we identified were $(x-10)$ and $(x-2)$. When multiplied together, these give us $x^2 - 12x + 20$. This directly matches Option A. The other options, when factored, resulted in different zeros. Option B didn't factor nicely into integers, and Options C and D yielded negative zeros.

Let's quickly double-check by plugging in the values into Option A:

• For $x=10$: $f(10) = (10)^2 - 12(10) + 20 = 100 - 120 + 20 = -20 + 20 = 0$. Correct!
• For $x=2$: $f(2) = (2)^2 - 12(2) + 20 = 4 - 24 + 20 = -20 + 20 = 0$. Correct!

So, there you have it! The function $f(x) = x^2 - 12x + 20$ definitively has zeros at $x=10$ and $x=2$. This whole process really highlights the power of factoring and understanding the relationship between zeros and factors of a polynomial. Keep practicing these, guys, and you'll be a math whiz in no time! Remember, understanding the core concepts makes solving these problems much easier and, dare I say, even fun!