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Chapter 2: Problem 3

Let \(g\) be the function defined by \(g(x)=3 x^{2}-6 x-3\). Find \(g(0), g(-1), g(a), g(-a)\), and \(g(x+1) .\)

Short Answer

Expert verified
The short answer is: \[g(0) = -3, \ g(-1) = 6, \ g(a) = 3a^2 - 6a - 3, \ g(-a) = 3a^2 + 6a - 3, \ g(x+1) = 3x^2\]

Step by step solution

01

Find g(0)

To find g(0), replace x with 0 in the given function, and calculate the result: \[g(0)=3(0)^{2}-6(0)-3\] This simplifies to: \[g(0)=-3\]
02

Find g(-1)

To find g(-1), replace x with -1 in the given function, and calculate the result: \[g(-1)=3(-1)^{2}-6(-1)-3\] This simplifies to: \[g(-1)=6\]
03

Find g(a)

To find g(a), replace x with a in the given function, and calculate the result: \[g(a)=3(a)^{2}-6(a)-3\] This cannot be further simplified, so we have: \[g(a) = 3a^2 - 6a - 3\]
04

Find g(-a)

To find g(-a), replace x with -a in the given function, and calculate the result: \[g(-a)=3(-a)^{2}-6(-a)-3\] This simplifies to: \[g(-a)=3a^2 + 6a - 3\]
05

Find g(x+1)

To find g(x+1), replace x with x+1 in the given function, and calculate the result: \[g(x+1)=3(x+1)^{2}-6(x+1)-3\] Now, expand and simplify the expression: \begin{align*} g(x+1) &= 3(x^2+2x+1)-6(x+1)-3 \\ &= 3x^2+6x+3-6x-6-3 \\ &= 3x^2 \end{align*} So, the final results are: \[g(0)=-3\] \[g(-1)=6\] \[g(a)=3a^2-6a-3\] \[g(-a)=3a^2+6a-3\] \[g(x+1)=3x^2\]

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Key Concepts

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

Function Evaluation
Function evaluation refers to the process of determining the output of a function for specific input values. It is fundamental to understanding how functions map inputs to outputs. To evaluate a function, you substitute the given input value into the variable present in the function's expression. This helps us see how the function behaves at specific points or under certain conditions.

Let's explore how this works using the function\( g(x) = 3x^2 - 6x - 3 \):
  • When evaluating \( g(0) \), substitute \( x = 0 \) into the equation to get \( g(0) = 3(0)^2 - 6(0) - 3 = -3 \).
  • For \( g(-1) \), substituting \( x = -1 \) yields \( g(-1) = 3(-1)^2 - 6(-1) - 3 = 6 \).
  • When finding \( g(a) \), use \( x = a \) for the substitution, giving \( g(a) = 3a^2 - 6a - 3 \).
Function evaluation is a straightforward yet essential skill. It allows you to explore the properties and graph of the function by obtaining specific points.
Quadratic Functions
Quadratic functions are polynomial functions of degree two. They take the standard form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These functions are represented by a parabola when graphed on a coordinate plane.

Our function, \( g(x) = 3x^2 - 6x - 3 \), is a quadratic function with:
  • Coefficient \( a = 3 \) governing the width and direction of the parabola (since \( a > 0 \), it opens upwards).
  • Coefficient \( b = -6 \) affecting the axis of symmetry and vertex location.
  • Constant term \( c = -3 \) representing the y-intercept, the point where the graph crosses the y-axis. Here, the intercept is at \( (0, -3) \).
Quadratic functions are vital in various applications, from physics to economics, due to their neatly structured form and predictable behavior. Understanding how to manipulate and graph these functions is a key algebra skill.
Algebraic Manipulation
Algebraic manipulation involves the rearrangement of expressions and solving equations to simplify or solve these expressions for desired variables. It is a core tool in algebra that helps transform functions and equations into a more usable or understandable form.

For instance, with \( g(x + 1) \), we begin by substituting \( x + 1 \) into \( g(x) \) and then simplify:
  • The initial substitution gives \( g(x + 1) = 3(x + 1)^2 - 6(x + 1) - 3 \).
  • Expanding \( (x + 1)^2 \) results in \( x^2 + 2x + 1 \).
  • When we substitute and simplify, we get \( g(x + 1) = 3(x^2 + 2x + 1) - 6(x + 1) - 3 = 3x^2 + 6x + 3 - 6x - 6 - 3 \).
  • Combining like terms results in \( g(x + 1) = 3x^2 \), a much simpler form.
Algebraic manipulation allows us to see relationships and patterns in functions more easily, enhancing understanding and problem-solving skills.

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Most popular questions from this chapter

Find the vertex, the \(x\) -intercepts (if any), and sketch the parabola. \(f(x)=1.2 x^{2}+3.2 x-1.2\)

According to a study conducted in 2004, the number of subscribers of BlackBerry, the handheld email devices manufactured by Research in Motion Ltd., is approximated by \(N(t)=-0.0675 t^{4}+0.5083 t^{3}-0.893 t^{2}+0.66 t+0.32\) \((0 \leq t \leq 4)\) where \(N(t)\) is measured in millions and \(t\) in years, with \(t=0\) corresponding to the beginning of 2002 . a. How many BlackBerry subscribers were there at the beginning of \(2002 ?\) b. How many BlackBerry subscribers were there at the beginning of \(2006 ?\)

The rise of digital music and the improvement to the DVD format are part of the reasons why the average selling price of standalone DVD recorders will drop in the coming years. The function $$ A(t)=\frac{699}{(t+1)^{0.94}} \quad(0 \leq t \leq 5) $$ gives the projected average selling price (in dollars) of standalone DVD recorders in year \(t\), where \(t=0\) corresponds to the beginning of 2002 . What was the average selling price of standalone DVD recorders at the beginning of \(2002 ?\) At the beginning of \(2007 ?\)

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. If the profit function is given by \(P(x)=a x^{2}+b x+c\), where \(x\) is the number of units produced and sold, then the level of production that yields a maximum profit is \(-\frac{b}{2 a}\) units.

Lynbrook West, an apartment complex, has 100 two-bedroom units. The monthly profit realized from renting out \(x\) apartments is given by $$ P(x)=-10 x^{2}+1760 x-50,000 $$ dollars. How many units should be rented out in order to maximize the monthly rental profit? What is the maximum monthly profit realizable?
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