Problem 40 - Intermediate Algebra

Question

If \(g(x)=x^{2}+3 x-1\), find \(g(1), g(-1), g(3)\), and \(g(-4)\).

Step-by-Step Solution

Answer

\(g(1) = 3\), \(g(-1) = -3\), \(g(3) = 17\), \(g(-4) = 3\).

Step 1: Substitute x with 1 in g(x)

To find \(g(1)\), substitute \(x = 1\) into the function \(g(x) = x^2 + 3x - 1\). This results in \(g(1) = 1^2 + 3 \times 1 - 1\).

Step 2: Calculate g(1)

Compute the expression from Step 1: \(g(1) = 1 + 3 - 1\). Now, calculate: \(1 + 3 = 4\), then subtracting 1 gives \(g(1) = 3\).

Step 3: Substitute x with -1 in g(x)

To find \(g(-1)\), substitute \(x = -1\) into the function: \(g(-1) = (-1)^2 + 3 \times (-1) - 1\).

Step 4: Calculate g(-1)

Compute the expression: \(g(-1) = 1 - 3 - 1\). Calculate: \(-3 - 1 = -4\), and adding 1 gives \(g(-1) = -3\).

Step 5: Substitute x with 3 in g(x)

To find \(g(3)\), substitute \(x = 3\) into the function: \(g(3) = 3^2 + 3 \times 3 - 1\).

Step 6: Calculate g(3)

Compute the expression: \(g(3) = 9 + 9 - 1\). Calculate: \(9 + 9 = 18\), then \(18 - 1 = 17\). So, \(g(3) = 17\).

Step 7: Substitute x with -4 in g(x)

To find \(g(-4)\), substitute \(x = -4\) into the function: \(g(-4) = (-4)^2 + 3 \times (-4) - 1\).

Step 8: Calculate g(-4)

Compute the expression: \(g(-4) = 16 - 12 - 1\). Calculate: \(-12 - 1 = -13\), and adding 16 gives \(g(-4) = 3\).

Key Concepts

Quadratic Functions

Quadratic functions are a type of polynomial that are characterized by the highest degree term being squared. In other words, they take the general form of \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. The function given in the exercise, \(g(x) = x^2 + 3x - 1\), is a perfect example of a quadratic function because it features an \(x^2\) term.

Quadratic functions form a parabola when plotted on a graph. This shape may open upwards or downwards, depending on the coefficient \(a\). In our function \(g(x)\), \(a = 1\), which means the parabola opens upwards. Key features of a parabola include the vertex, which is the highest or lowest point, and the axis of symmetry, which is a vertical line that passes through the vertex.

The solutions or roots of a quadratic function are the values of \(x\) for which the function equals zero. These can be found using various methods including factoring, completing the square, or using the quadratic formula. However, in this particular exercise, we are focused on function evaluation rather than finding roots.

Polynomial Evaluation

Polynomial evaluation involves finding the output of a polynomial function for specific values of the variable. For example, in the exercise, we are asked to evaluate the function \(g(x) = x^2 + 3x - 1\) at several points, namely \(x = 1, -1, 3,\) and \(-4\).

This is done by substituting the given values for \(x\) into the function and simplifying the expression to calculate the result. This process helps us understand how the function behaves at different points, and it can be useful for graphing or determining specific points of interest, such as intercepts.

Upon substituting these values, we can calculate as follows:

• For \(g(1)\): Plug \(1\) into the function to get \(1^2 + 3 \times 1 - 1 = 3\).
• For \(g(-1)\): Plug \(-1\) into the function to get \((-1)^2 + 3 \times (-1) - 1 = -3\).
• For \(g(3)\): Plug \(3\) into the function to get \(3^2 + 3 \times 3 - 1 = 17\).
• For \(g(-4)\): Plug \(-4\) into the function to get \((-4)^2 + 3 \times (-4) - 1 = 3\).

By using these substitutions, you can directly evaluate polynomials at any point.

Algebraic Substitution

Algebraic substitution is a fundamental technique in algebra that allows us to replace a variable in an expression with a given number or another expression. This method is essential for evaluating functions, simplifying expressions, and solving equations.

Take the given function \(g(x) = x^2 + 3x - 1\) as an example. We substitute different values for \(x\) to find specific outputs, such as \(g(1)\), \(g(-1)\), \(g(3)\), and \(g(-4)\). Here's how it works in detail:

• Start by replacing every instance of \(x\) in the function with the number you are evaluating.
• Then, perform the necessary arithmetic operations: multiply, add, and subtract as required.
• Simplify the expression step-by-step until reaching the final result.

This procedure of substitution followed by simplification helps to efficiently compute the function's value for given inputs.

Mastering this skill can aid in working with much more complex functions and is a building block for calculus and other advanced mathematical concepts. Practicing substitution will improve your ability to solve a wide range of mathematical problems.