/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Let \(f(x)=2 x^{2}+7 x+5\). Eval... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 8

Let \(f(x)=2 x^{2}+7 x+5\). Evaluate (a) \(f(3)\) (b) \(f(a)\) (c) \(f(2 a)\) (d) \(f(-2)\)

Short Answer

Expert verified
Answer: The values of the function are: (a) \(f(3) = 44\) (b) \(f(a) = 2a^2 + 7a + 5\) (c) \(f(2a) = 8a^2 + 14a + 5\) (d) \(f(-2) = -1\)

Step by step solution

01

(a) Evaluate f(3)

To evaluate the function at \(x=3\), we will replace the variable x with 3: \(f(3) = 2(3)^2 + 7(3) + 5\) Now, we will calculate the value: \(f(3) = 2(9) + 21 + 5 \) \(f(3) = 18 + 21 + 5 \) \(f(3) = 44\)
02

(b) Evaluate f(a)

To evaluate the function at \(x=a\), we will replace the variable x with 'a': \(f(a) = 2(a)^2 + 7(a) + 5\) There is nothing more to calculate in this case, so our answer is: \(f(a) = 2a^2 + 7a + 5\)
03

(c) Evaluate f(2a)

To evaluate the function at \(x=2a\), we will replace the variable x with '2a': \(f(2a) = 2(2a)^2 + 7(2a) + 5\) Now, we will simplify the expression: \(f(2a) = 2(4a^2) + 14a + 5 \) \(f(2a) = 8a^2 + 14a + 5\)
04

(d) Evaluate f(-2)

To evaluate the function at \(x=-2\), we will replace the variable x with '-2': \(f(-2) = 2(-2)^2 + 7(-2) + 5\) Now, we will calculate the value: \(f(-2) = 2(4) - 14 + 5 \) \(f(-2) = 8 - 14 + 5 \) \(f(-2) = -1\) So we have: (a) \(f(3) = 44\) (b) \(f(a) = 2a^2 + 7a + 5\) (c) \(f(2a) = 8a^2 + 14a + 5\) (d) \(f(-2) = -1\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Functions
A quadratic function is a special type of polynomial function where the highest degree of the variable is 2. This means the variable, usually represented as \(x\), is squared. The standard form of a quadratic function is \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). The graph of a quadratic function is a parabola, which can either open upwards or downwards depending on the sign of \(a\).
  • An upward opening parabola occurs when \(a > 0\).
  • A downward opening parabola occurs when \(a < 0\).
In the exercise above, the quadratic function \(f(x) = 2x^2 + 7x + 5\) has \(a = 2\), \(b = 7\), and \(c = 5\).
This function will open upwards since \(a\) is positive.
Function Notation
Function notation is a way to name and define functions in terms of their inputs and outputs. It is expressed in the form \(f(x)\), where \(f\) represents the function and \(x\) is the variable or input of the function. The output of the function corresponds to a particular input plugged into the function's formula.

For example, in the function \(f(x) = 2x^2 + 7x + 5\), \(f\) represents the function and the expression \(2x^2 + 7x + 5\) is how it transforms inputs.
The function notation helps evaluate the function at specific values:
  • To find \(f(3)\), you substitute \(3\) into the function, replacing all instances of \(x\).
  • To find \(f(a)\), you replace \(x\) with the variable \(a\).
  • It allows us to explore a variety of scenarios by substituting different values or expressions in place of \(x\).
This elegance and simplicity make function notation a powerful tool in mathematics.
Polynomial Evaluation
Polynomial evaluation involves substituting a specific number or expression into a polynomial function to calculate the resulting value or simplified expression. For instance, with \(f(x) = 2x^2 + 7x + 5\), evaluating the polynomial at \(x = 3\), \(x = a\), \(x = 2a\), and \(x = -2\) involves replacing \(x\) by these numbers or expressions. Here is how polynomial evaluation works:
  • Direct Substitution: Replace \(x\) with the given value or expression, like substituting \(3\) for \(x\) to find \(f(3)\).
  • Simplification: After replacing \(x\), simplify the expression following the order of operations: exponents, multiplication and division, then addition and subtraction.
  • Adaptive Use: This method allows us to handle diverse inputs and find the output efficiently. Using \(f(2a)\) in the place of \(f(x)\), we adjust operations to suit different inputs.
Polynomial evaluation is a fundamental skill for solving algebraic expressions and understanding more complex mathematical concepts.

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