Problem 71 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the zeros (if any) of the rational function. $$f(x)=\frac{x^{2}-9}{x+1}$$

Step-by-Step Solution

Answer

The zeros of the function \(f(x)=\frac{x^{2}-9}{x+1}\) are \(x = 3\) and \(x = -3\).

Step 1: Solve the numerator

To find the potential zeros of \(f(x)\), we begin with setting the numerator \(x^{2}-9\) equal to zero and solve for \(x\): \(x^{2}-9 = 0\) This equation is a difference of squares. We can write \(x^{2}-9\) as \((x-3)(x+3)\) which helps us solve for \(x\). \((x-3)(x+3) = 0\) Therefore, \(x = 3\) or \(x = -3\).

Step 2: Check the denominator

Next, we need to make sure that \(x = 3\) and \(x = -3\) do not make the denominator equal to zero: If \(x = 3\), the denominator equals \(3 + 1 = 4\). If \(x = -3\), the denominator equals \(-3 + 1 = -2\). So neither of these values makes the denominator zero.

Step 3: Conclusion

The zeros of the function \(f(x)=\frac{x^{2}-9}{x+1}\) are \(x = 3\) and \(x = -3\), since these values make the function equal to zero and do not make the denominator zero.

Key Concepts

Zeros of a Function

When we talk about the "zeros" of a function, it means we're looking for the values of the variable (often represented as \(x\)) where the function equals zero. For a rational function, this is primarily determined by the numerator of the function. If the numerator is zero while the denominator is not zero, the function itself becomes zero at those points. In this case, if we have the function \(f(x) = \frac{x^2 - 9}{x + 1}\), we'll find the zeros by setting the numerator \(x^2 - 9\) equal to zero and solving for \(x\).

• Set \(x^2 - 9 = 0\).
• Solve the equation to find the values of \(x\).
• This leads to two potential zeros: \(x = 3\) and \(x = -3\).

It's crucial to verify that these values do not also make the denominator zero, as that would make them invalid as zeros for the function.

Difference of Squares

The term "difference of squares" refers to a specific algebraic expression that takes the form \(a^2 - b^2\). This expression can be factored into two binomials: \((a-b)(a+b)\). This factorization is very handy in solving polynomial equations because it simplifies the expression into factors that are much easier to solve.For the function \(f(x) = \frac{x^2 - 9}{x + 1}\), the numerator \(x^2 - 9\) is a perfect example of a difference of squares: here, \(a = x\) and \(b = 3\). By using the difference of squares method, we can factor the numerator as:

• \(x^2 - 9 = (x - 3)(x + 3)\)

This separation into \((x - 3)\) and \((x + 3)\) makes solving the equation \(x^2 - 9 = 0\) straightforward, as we can directly determine that \(x = 3\) or \(x = -3\).

Numerator and Denominator in Fractions

In rational functions, the numerator and the denominator play crucial roles. The function is expressed as a fraction: \(\frac{P(x)}{Q(x)}\), where \(P(x)\) is the numerator and \(Q(x)\) is the denominator. The numerator specifies where the function could potentially cross the x-axis (those are the zeros), while the denominator determines the values that \(x\) must not take, as they would cause division by zero.In our example, \(f(x) = \frac{x^2 - 9}{x + 1}\), let's identify:

• Numerator: \(x^2 - 9\)
• Denominator: \(x + 1\)

The zeros of \(f(x)\) are found by setting the numerator \(x^2 - 9\) to zero, yielding values \(x = 3\) and \(x = -3\). However, you must always check these values against the denominator because if the denominator also equals zero at these points, the function would not be defined there. For \(x + 1 = 0\), the zero point would be \(x = -1\). Hence, in our given function, at \(x = 3\) and \(x = -3\), the denominator does not equal zero, confirming their validity as zeros.