Chapter 4: Problem 248
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.
Short Answer
Expert verified
The x-intercepts are at x = -1 and x = -3.
Step by step solution
01
Set Function to Zero
To find the \(x\)-intercepts, we need to determine when the function \(f(x) = \log(x^2 + 4x + 4)\) is equal to zero. This is because the \(x\)-intercepts are the points where the graph crosses the \(x\)-axis, implying \(f(x) = 0\). So we start with the equation \(\log(x^2 + 4x + 4) = 0\).
02
Solve the Logarithmic Equation
To solve \(\log(x^2 + 4x + 4) = 0\), we convert the logarithmic equation into an exponential form. The equation \(\log_a(b) = c\) implies that \(b = a^c\). Here, assuming the log is base 10, we have \(x^2 + 4x + 4 = 10^0 = 1\).
03
Solve the Quadratic Equation
Now solve the quadratic equation \(x^2 + 4x + 4 = 1\). This simplifies to \(x^2 + 4x + 3 = 0\). To solve it, we factor the quadratic as \((x + 1)(x + 3) = 0\).
04
Find the Roots
The factored equation \((x + 1)(x + 3) = 0\) gives us the solutions by setting each factor equal to zero: \(x + 1 = 0\) or \(x + 3 = 0\). Solving these gives \(x = -1\) and \(x = -3\).
05
Verify by Graphing
Upon graphing \(f(x) = \log(x^2 + 4x + 4)\), we observe that the \(x\)-intercepts, where the graph crosses the \(x\)-axis, are indeed at \(x = -1\) and \(x = -3\). This confirms our algebraic solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithmic Functions
Logarithmic functions are an essential concept in algebra and calculus. A logarithm answers the question, "To what exponent must we raise a base, usually 10 or 'e', to obtain a given number?" For example, the mathematical expression \( \log_{10}(x) \) essentially translates to "the power to which you must raise 10 to get x." In the exercise, we use a logarithmic function to find the x-intercepts of \(f(x) = \log(x^2 + 4x + 4)\). Understanding logarithms involves knowing certain properties, such as:
- \( \log_a(1) = 0 \), because any number raised to the zero power is 1.
- \( \log_a(a) = 1 \), since a to the first power is a.
- Conversion to exponential form: \( \log_a(b) = c \Rightarrow b = a^c \)
Quadratic Equations
Quadratic equations often appear in algebra, typically in the standard form \(ax^2 + bx + c = 0\). They represent parabolic curves when graphed. In the exercise, after converting the logarithmic equation, we end up with the quadratic equation \(x^2 + 4x + 3 = 0\). Solving quadratic equations can be done in several ways:
- Factoring: Expressing the quadratic as a product of binomials, like \((x + 1)(x + 3) = 0\).
- Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
- Completing the square: Rewriting the quadratic in the form \((x - p)^2 = q\).
X-Intercepts
Finding x-intercepts is crucial when analyzing the graph of a function. An x-intercept of a function is a point where the graph crosses the x-axis, meaning the function value is zero there (\(f(x) = 0\)). In our exercise, we found the x-intercepts algebraically by setting \(\log(x^2 + 4x + 4) = 0\) and solving the subsequent equations. The solutions to these equations, \(x = -1\) and \(x = -3\), tell us where the function cuts the x-axis. It’s important because it gives us key insight into the roots of the function and can assist in sketching more accurate graphs.
Graphing Functions
Graphing functions is a visual method to understand and analyze the behavior of mathematical expressions. In our solution, graphing \(f(x) = \log(x^2 + 4x + 4)\) helped verify the results found algebraically. To graph a function like this:
- Find and plot significant points: x-intercepts, y-intercepts, and any other calculated points where the function may change direction.
- Determine the domain and range to know which x and y values make sense for your graph.
- Look for symmetry, asymptotes, or transformations that can simplify your graphing process.