Problem 70 - College Algebra with Corequisite Support

Question

For the following exercises, evaluate the function \(f\) at the values \(f(-2), f(-1), f(0), f(1),\) and \(f(2) .\) \(f(x)=8 x^{2}-7 x+3\)

Step-by-Step Solution

Answer

\(f(-2) = 49, f(-1) = 18, f(0) = 3, f(1) = 4, f(2) = 21\).

Step 1: Substitute -2 into the function

Let's evaluate the function at \(x = -2\). Replace each instance of \(x\) in \(f(x) = 8x^2 - 7x + 3\) with \(-2\). This becomes:\[f(-2) = 8(-2)^2 - 7(-2) + 3\]

Step 2: Calculate the function at x = -2

Calculate \(8(-2)^2 = 8 \times 4 = 32\), \(-7(-2) = 14\), and add 3. So:\[f(-2) = 32 + 14 + 3 = 49\]

Step 3: Substitute -1 into the function

Now, evaluate the function at \(x = -1\). Replace \(x\) with \(-1\):\[f(-1) = 8(-1)^2 - 7(-1) + 3\]

Step 4: Calculate the function at x = -1

Calculate \(8(-1)^2 = 8\), \(-7(-1) = 7\), and add 3. So:\[f(-1) = 8 + 7 + 3 = 18\]

Step 5: Substitute 0 into the function

Substitute \(0\) into the function:\[f(0) = 8(0)^2 - 7(0) + 3\]

Step 6: Calculate the function at x = 0

Since any term with \(x\) becomes zero, we only have:\[f(0) = 3\]

Step 7: Substitute 1 into the function

Next, evaluate the function at \(x = 1\). Substitute \(x\) with \(1\):\[f(1) = 8(1)^2 - 7(1) + 3\]

Step 8: Calculate the function at x = 1

Calculate \(8(1)^2 = 8\), \(-7(1) = -7\), and add 3. So:\[f(1) = 8 - 7 + 3 = 4\]

Step 9: Substitute 2 into the function

Finally, evaluate the function at \(x = 2\). Substitute \(x\) with \(2\):\[f(2) = 8(2)^2 - 7(2) + 3\]

Step 10: Calculate the function at x = 2

Calculate \(8(2)^2 = 32\), \(-7(2) = -14\), and add 3. So:\[f(2) = 32 - 14 + 3 = 21\]

Key Concepts

Function Evaluation

Function evaluation is the process of determining the value of a function for specific values of its variables. Imagine you have a function, which is like a machine. You feed the machine with certain values and it gives you a result back. In mathematical terms, given a function \(f(x)\), you replace \(x\) with the value of interest to find out what's the outcome.To evaluate a function:

• Identify the function expression, like \(f(x) = 8x^2 - 7x + 3\).
• Substitute the values you're interested in, one at a time, for \(x\). For instance, to find \(f(-2)\), plug \(-2\) into the function wherever you see \(x\).
• Perform the necessary arithmetic operations to simplify the expression and arrive at the result.

Approach each evaluation systematically and remember that practice enhances your understanding.

Polynomial Expressions

A polynomial expression is an algebraic expression made up of terms, each consisting of a variable raised to an exponent and multiplied by a coefficient. In its simplest form, it's like a sentence in algebra, involving the addition, subtraction, and multiplication of variables and constants.With the given function \(f(x) = 8x^2 - 7x + 3\), you can observe its components:

• The term \(8x^2\) is a quadratic term, as the variable \(x\) is raised to the power 2.
• The term \(-7x\) is the linear term, where \(x\) is raised to the power 1.
• The term \(+3\) is the constant term. It remains unchanged regardless of \(x\)'s value.

Understanding polynomial expressions helps in grasping how different inputs yield various outputs, crucial for evaluating functions and finding solutions to equations.

Substitution Method

The substitution method is a technique used to evaluate expressions, especially helpful with polynomials. The idea is to replace variables with specific values and simplify the expression to find an answer.Here's how you can do substitution effectively:

• Start by clearly identifying the variable to be replaced. In our function \(f(x) = 8x^2 - 7x + 3\), \(x\) is replaced by values like -2, -1, 0, 1, and 2.
• Everywhere you see the variable in the expression, substitute it with the given value.
• Perform arithmetic calculations carefully. Simplify using BIDMAS/BODMAS rules: brackets, indices (powers), division/multiplication, addition/subtraction.

Through substitution, complex expressions become simpler, allowing straightforward arithmetic to solve otherwise complicated problems.