Problem 58
Question
Find \(f(x)\) at the indicated value of \(x\). $$f(x)=-x^{2}-x-6, x=3$$
Step-by-Step Solution
Verified Answer
\( f(3) = -18 \)
1Step 1: Understand the Problem
The task is to find the value of the function \( f(x) = -x^2 - x - 6 \) when \( x = 3 \). This means we substitute \( x = 3 \) into the function and compute the resulting value.
2Step 2: Substitute the Value
Replace \( x \) in the function with 3: \( f(3) = -(3)^2 - (3) - 6 \). This substitution will allow us to calculate the exact value of the function at \( x = 3 \).
3Step 3: Simplify the Expression
Calculate \( (3)^2 \) which equals 9. Then compute \( -9 - 3 - 6 \). This simplifies to \( f(3) = -9 - 3 - 6 \).
4Step 4: Compute the Result
First compute \( -9 - 3 = -12 \), then \( -12 - 6 = -18 \). Therefore, \( f(3) = -18 \).
Key Concepts
Quadratic FunctionSubstitutionSimplification Steps
Quadratic Function
A quadratic function is essentially any polynomial function where the highest degree of the variable is 2. Such functions take the general form of \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero. Here, the variable \( x \) is raised to the power of 2, making it a 'quadratic' term.
Quadratic functions are easily recognizable by their unique U-shaped curve when graphed, known as a parabola. The direction the parabola opens (upward or downward) is determined by the sign of the coefficient \( a \). If \( a \) is positive, the parabola opens upward, and if negative, it opens downward. In our example, \( f(x) = -x^2 - x - 6 \), the coefficient of \( x^2 \) is negative, which means the parabola will open downwards. This particular function has a few key components:
Quadratic functions are easily recognizable by their unique U-shaped curve when graphed, known as a parabola. The direction the parabola opens (upward or downward) is determined by the sign of the coefficient \( a \). If \( a \) is positive, the parabola opens upward, and if negative, it opens downward. In our example, \( f(x) = -x^2 - x - 6 \), the coefficient of \( x^2 \) is negative, which means the parabola will open downwards. This particular function has a few key components:
- Leading coefficient: The \( x^2 \) term, which is \(-1\) in this case, defines the width and direction of the parabola.
- Linear term: The \( x \) term, which is \(-1\), affects the slope and offset of the parabola.
- Constant term: The number \(-6\) serves as the y-intercept, the point where the parabola crosses the y-axis.
Substitution
Substitution is a fundamental step in function evaluation, especially crucial in simplifying expressions and getting a numerical result. When substituting, you simply replace every instance of the variable in the function with a specified number.
In the problem we're exploring, we need to find the value of \( f(x) = -x^2 - x - 6 \) when \( x = 3 \). This is done by substituting \( x = 3 \) everywhere in the equation, transforming it as follows:
In the problem we're exploring, we need to find the value of \( f(x) = -x^2 - x - 6 \) when \( x = 3 \). This is done by substituting \( x = 3 \) everywhere in the equation, transforming it as follows:
- Start with the original expression: \( f(x) = -x^2 - x - 6 \).
- Replace every \( x \) with 3: \( f(3) = -(3)^2 - 3 - 6 \).
Simplification Steps
Simplification is the process of converting expressions into their simplest possible form, often by performing arithmetic operations. This stage is crucial in determining the evaluated value of a function after substitution.
Once the substitution is done, as in \( f(3) = -(3)^2 - 3 - 6 \), the following simplification steps should be undertaken:
Once the substitution is done, as in \( f(3) = -(3)^2 - 3 - 6 \), the following simplification steps should be undertaken:
- Compute the square of 3: \( (3)^2 = 9 \).
- Substitute this back into the expression: \( f(3) = -9 - 3 - 6 \).
- Proceed with arithmetic operations from left to right: \(-9 - 3 = -12\).
- Continue simplifying to add the last term: \(-12 - 6 = -18\).
Other exercises in this chapter
Problem 57
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