Problem 62 - College Algebra

Question

Use properties of exponents to find the \(x\) -intercepts of the function \(f(x)=\log \left(x^{2}+4 x+4\right)\) algebraically. Show the steps for solving, and then verify the result by graphing the function.

Step-by-Step Solution

Answer

The \(x\)-intercepts are \(x = -1\) and \(x = -3\).

Step 1: Understand the Problem

The function given is \( f(x) = \log(x^2 + 4x + 4) \). The \(x\)-intercepts are the values of \(x\) for which \(f(x) = 0\). So we need to solve the equation \(\log(x^2 + 4x + 4) = 0\).

Step 2: Set the Logarithm Equal to Zero

For the logarithm \(\log(y) = 0\), the argument \(y\) must equal 1. Hence for \(\log(x^2 + 4x + 4) = 0\), we set \(x^2 + 4x + 4 = 1\).

Step 3: Solve the Quadratic Equation

By setting \(x^2 + 4x + 4 = 1\), we rearrange to obtain \(x^2 + 4x + 3 = 0\). This is a quadratic equation. To solve it, we use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 4\), and \(c = 3\).

Step 4: Apply the Quadratic Formula

Substituting into the formula, we get:\[x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 - 12}}{2} = \frac{-4 \pm \sqrt{4}}{2} = \frac{-4 \pm 2}{2}\]Thus, the solutions are \(x = \frac{-4 + 2}{2} = -1\) and \(x = \frac{-4 - 2}{2} = -3\).

Step 5: Verify the Solutions

These solutions \(x = -1\) and \(x = -3\) can be verified by graphing the function \(f(x)\). Both should be \(x\)-intercepts, which means they should appear as points where the graph crosses the x-axis.

Key Concepts

Properties of Exponents

Understanding the properties of exponents is essential in various algebraic problems, especially when they involve logarithmic and exponential functions. Exponential properties help us simplify and solve equations that involve powers. Some basic properties are:

• Product of Powers: When multiplying two exponents with the same base, add the exponents. That means, \( a^m \cdot a^n = a^{m+n} \).
• Quotient of Powers: When dividing two exponents with the same base, subtract the exponent in the denominator from the exponent in the numerator, \( \frac{a^m}{a^n} = a^{m-n} \).
• Power of a Power: When an exponent is raised to another exponent, multiply the exponents, \( (a^m)^n = a^{m\cdot n} \).

These rules simplify the processes involved in solving algebraic equations that have powers, such as rearranging equations to solve for variables.

X-intercepts

In an algebraic context, x-intercepts are the real values of \(x\) where a function crosses the x-axis, meaning the value of the function is zero at these points. To find x-intercepts algebraically in a function \(f(x)\), set \(f(x) = 0\) and solve for \(x\). This calculation transforms the problem into a typical equation-solving scenario often involving factoring, using the quadratic formula, or other algebraic methods.In the given exercise, having the function \(f(x) = \log(x^2 + 4x + 4)\), we're interested in finding where this expression equals zero. This entails solving \(x^2 + 4x + 4 = 1\) to identify the x-intercepts.

Quadratic Formula

The quadratic formula is a crucial tool when dealing with quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). The formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) provides the solution directly by substituting the coefficients \(a\), \(b\), and \(c\). Here’s how to use it:

• Identify the coefficients: In any quadratic equation, ensure you correctly spot \(a\), \(b\), and \(c\). For instance, in \(x^2 + 4x + 3 = 0\), \(a = 1\), \(b = 4\), and \(c = 3\).
• Inside the formula: Calculate the discriminant \(b^2 - 4ac\), which determines the number of real solutions.
• Substitute into the formula: Insert the values into \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to get solutions.

In our exercise, applying this formula solved \(x^2 + 4x + 3 = 0\) and gave solutions \(x = -1\) and \(x = -3\). These are the x-intercepts of the function.

Solving Equations

Solving equations is a fundamental skill in algebra, focusing on finding a value for the variable that makes the equation true. In the context of quadratic equations, solving often involves transforming the original equation into a standard form and applying methods like factoring, completing the square, or using the quadratic formula. Here's a general approach to solving quadratic equations:

• Standard Form: Make sure the equation is in the form \(ax^2 + bx + c = 0\).
• Choosing a Method: If the equation is easily factorable, factoring could provide quick solutions. Use the quadratic formula if factoring is complex or for checking.
• Checking the Solutions: Plug the solutions back into the original equation to verify they work, especially considering they should satisfy any other condition provided in the problem (e.g., in logarithmic contexts, ensuring the argument is positive).

In our specific problem, we applied the quadratic formula to the reduced equation from the logarithmic setting and verified our solutions by checking they fulfilled the original equation's conditions.