Find The Function Equation From Its Zeros

When we talk about the zeros of a function, we're essentially looking for the input values (often represented as 'x') that make the function's output (often represented as 'f(x)' or 'y') equal to zero. Think of it as finding where the graph of the function crosses or touches the x-axis. If a function $f$ has zeros at $x = a$ and $x = b$, it means that $f(a) = 0$ and $f(b) = 0$. A common way to represent a function with known zeros is by using its factored form. If the zeros are $a$ and $b$, then the function can be expressed as $f(x) = k(x-a)(x-b)$, where $k$ is a non-zero constant. The constant $k$ allows for vertical stretching or compression of the graph and can also represent a reflection across the x-axis if $k$ is negative. However, when we are asked to find an equation that could represent the function, and we are given the zeros, we can often assume $k=1$ for simplicity, unless there's other information provided, like a specific point the function passes through. This is because many possible equations could have the same zeros, differing only by this constant factor $k$. The core idea is that the factors $(x-a)$ and $(x-b)$ must be present in the equation so that when $x=a$ or $x=b$, the corresponding factor becomes zero, making the entire product zero. For instance, if a function has a zero at $x = -1$, then $(x - (-1))$, which simplifies to $(x+1)$, must be a factor of the function. Similarly, if a function has a zero at $x = -5$, then $(x - (-5))$, which simplifies to $(x+5)$, must be another factor. Therefore, a function with zeros at -1 and -5 can be represented by an equation that includes the factors $(x+1)$ and $(x+5)$. When these factors are multiplied together, $(x+1)(x+5)$, we get $x^2 + 5x + x + 5$, which simplifies to $x^2 + 6x + 5$. So, a possible representation of the function $f$ could be $f(x) = (x+1)(x+5)$, or more generally, $f(x) = k(x+1)(x+5)$ for any non-zero constant $k$. This understanding is fundamental in algebra when working with polynomials and their roots.

Let's delve deeper into how zeros relate to the factored form of a polynomial function. The fundamental theorem of algebra states that a polynomial of degree $n$ has exactly $n$ complex roots (counting multiplicity). For a polynomial function $f(x)$, its zeros are precisely the roots of the equation $f(x) = 0$. If we are given that the zeros of a function $f$ are $-1$ and $-5$, this means that when we substitute $-1$ for $x$ in the function, the result is $0$, and when we substitute $-5$ for $x$, the result is also $0$. In mathematical terms, $f(-1) = 0$ and $f(-5) = 0$. The factor theorem is a crucial concept here. It states that a polynomial $f(x)$ has a factor $(x-c)$ if and only if $f(c) = 0$. In simpler words, if $c$ is a zero of the function, then $(x-c)$ is a factor of the polynomial. Applying this theorem to our problem: since $-1$ is a zero, $(x - (-1))$, which simplifies to $(x+1)$, must be a factor of $f(x)$. Likewise, since $-5$ is a zero, $(x - (-5))$, which simplifies to $(x+5)$, must also be a factor of $f(x)$. Therefore, the function $f(x)$ must contain both $(x+1)$ and $(x+5)$ as factors. The simplest form of such a function would be the product of these factors, possibly multiplied by a constant $k$. So, we can write $f(x) = k(x+1)(x+5)$. When we expand this, we get $f(x) = k(x^2 + 5x + x + 5) = k(x^2 + 6x + 5)$. For the purpose of selecting a possible equation from the given options, we can assume $k=1$, as this provides the simplest representation. This leads us to the equation $f(x) = (x+1)(x+5)$. Let's examine why the other options are incorrect based on this principle. If $f(x) = (x-1)(x+5)$, the zeros would be $x=1$ and $x=-5$. If $f(x) = (x-1)(x-5)$, the zeros would be $x=1$ and $x=5$. If $f(x) = (x+1)(x-5)$, the zeros would be $x=-1$ and $x=5$. Only the equation $f(x) = (x+1)(x+5)$ yields the given zeros of $-1$ and $-5$. This demonstrates the direct relationship between the zeros of a function and its factored form.

To solidify our understanding, let's consider the provided options and verify which one aligns with the given zeros: $-1$ and $-5$. The problem states that these are the zeros of the function $f$. This means that $f(-1) = 0$ and $f(-5) = 0$. We are looking for an equation that, when we substitute these values for $x$, results in $0$. Let's test each option:

Option 1: $f(x) = (x-1)(x+5)$ If we plug in $x = -1$, we get $f(-1) = (-1 - 1)(-1 + 5) = (-2)(4) = -8$. Since $f(-1) eq 0$, this is not the correct equation.

Option 2: $f(x) = (x-1)(x-5)$ If we plug in $x = -1$, we get $f(-1) = (-1 - 1)(-1 - 5) = (-2)(-6) = 12$. Since $f(-1) eq 0$, this is not the correct equation. If we plug in $x = -5$, we get $f(-5) = (-5 - 1)(-5 - 5) = (-6)(-10) = 60$. Again, not zero.

Option 3: $f(x) = (x+1)(x+5)$ Let's test $x = -1$: $f(-1) = (-1 + 1)(-1 + 5) = (0)(4) = 0$. This works! Now let's test $x = -5$: $f(-5) = (-5 + 1)(-5 + 5) = (-4)(0) = 0$. This also works! Since both $-1$ and $-5$ are zeros for this equation, this is a potential answer.

Option 4: $f(x) = (x+1)(x-5)$ If we plug in $x = -1$, we get $f(-1) = (-1 + 1)(-1 - 5) = (0)(-6) = 0$. This works. However, let's check $x = -5$: $f(-5) = (-5 + 1)(-5 - 5) = (-4)(-10) = 40$. Since $f(-5) eq 0$, this is not the correct equation.

Based on this thorough check, the only equation that correctly represents a function with zeros at $-1$ and $-5$ is $f(x) = (x+1)(x+5)$. This confirms our understanding of the relationship between zeros and the factored form of a function. The graph of this function would intersect the x-axis at the points $(-1, 0)$ and $(-5, 0)$. Expanding this equation gives us $f(x) = x^2 + 6x + 5$, which is a quadratic function. Understanding how to find the equation from the zeros is a fundamental skill in algebra, particularly when dealing with polynomials. It allows us to reconstruct functions based on their key characteristics, which is invaluable in various mathematical and scientific applications, from modeling physical phenomena to analyzing data. The ability to translate between zeros, factors, and the expanded form of a function is a cornerstone of working with functions. Remember that if a problem asked for all possible equations, we would need to consider the constant multiplier $k$, leading to $f(x) = k(x+1)(x+5)$. But since it asks for which equation could represent the function, the simplest form with $k=1$ is sufficient and usually the intended answer in such multiple-choice scenarios.

Conclusion

In summary, the zeros of a function are the values of $x$ for which $f(x) = 0$. According to the factor theorem, if $c$ is a zero of a function, then $(x-c)$ is a factor of that function. Given that the zeros of function $f$ are $-1$ and $-5$, we know that $(x - (-1))$, which is $(x+1)$, and $(x - (-5))$, which is $(x+5)$, must be factors of $f(x)$. Therefore, a possible representation of function $f$ is the product of these factors: $f(x) = (x+1)(x+5)$. By testing each option, we confirmed that only $f(x) = (x+1)(x+5)$ yields $f(-1)=0$ and $f(-5)=0$. This mathematical principle is a key concept in understanding polynomial functions and their properties. For further exploration into functions and their properties, you can visit Khan Academy's mathematics section.