Problem 18
Question
Decide whether each function is one-to-one. $$y=-(x+3)^{2}-8$$
Step-by-Step Solution
Verified Answer
The function is not one-to-one because it is a downward-opening quadratic.
1Step 1: Understanding One-to-One Functions
A function is considered one-to-one if every element of the range is mapped to by exactly one element of the domain. This means that if \( f(a) = f(b) \), then \( a = b \) must be true for the function to be one-to-one.
2Step 2: Analyzing the Given Function
The given function is \( y = -(x+3)^2 - 8 \). This is a quadratic function written in vertex form \( y = a(x-h)^2 + k \) with \( a = -1 \), \( h = -3 \), and \( k = -8 \). Since \( a < 0 \), the parabola opens downwards.
3Step 3: Testing for One-to-One Property
For a function to be one-to-one, each \( y \)-value should be unique for each \( x \)-value within its domain. Since the function is a downward-opening quadratic, it fails the horizontal line test (a line can intersect the parabola at more than one point), meaning it is not one-to-one.
Key Concepts
Quadratic FunctionsVertex FormHorizontal Line Test
Quadratic Functions
Quadratic functions are a type of polynomial function that are characterized by the highest degree term being squared. Their general form is given by \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These functions graph into a U-shaped curve known as a parabola. Here are some key features of quadratic functions:
- Vertex: The point where the parabola changes direction. For upward opening parabolas, it's the minimum point, and for downward opening ones, it's the maximum point.
- Axis of Symmetry: A vertical line that passes through the vertex, dividing the parabola into two mirror images.
- Direction of Opening: Determined by the sign of \( a \). If \( a > 0 \), the parabola opens upwards; if \( a < 0 \), it opens downwards.
Vertex Form
The vertex form of a quadratic function is a special representation that makes it easy to identify the vertex of the parabola. It is written as \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola. For our function, \( y = -(x+3)^2 - 8 \), the vertex is located at \((-3, -8)\). Understanding vertex form is crucial as it allows for a quick read of the parabola's key characteristics:
- Vertex: Easily identified as \((h, k)\) in the expression.
- Axis of Symmetry: The line \( x = h \) goes through the vertex and divides the parabola symmetrically.
- Direction: Determined by \( a \). Here, \( a = -1 \), so the parabola opens downward, indicating \( h \) is a maximum point.
Horizontal Line Test
The horizontal line test is a handy tool to determine if a function is one-to-one. A function is considered one-to-one if any horizontal line crosses its graph at most once. This test is essential when considering inverse functions since only one-to-one functions have inverses. In the context of quadratic functions, particularly our function \( y = -(x+3)^2 - 8 \), applying the horizontal line test reveals that the function is not one-to-one. A horizontal line can intersect the downward-opening parabola in more than one location, confirming that multiple \( x \)-values can result in the same \( y \)-value. In functions like this quadratic, the repeated intersection indicates many-to-one mapping within its domain, meaning not every output is unique. Thus, without restricting its domain, this particular quadratic cannot have an inverse function.
Other exercises in this chapter
Problem 17
March the logarithm in Column I with its value in Column II. Remember that \(\log _{a} x\) is the exponent to which a must be raised in onder to obtain \(x\). \
View solution Problem 17
Solve each exponential equation. Express the solution set so that (a) solutions are in exact form and, if irrational, (b) solutions are approximated to the near
View solution Problem 18
Find the domain of each logarithmic function analytically. You may wish to support your answer graphically. $$f(x)=\log _{6}\left(2 x^{2}-7 x-4\right)$$
View solution Problem 18
March the logarithm in Column I with its value in Column II. Remember that \(\log _{a} x\) is the exponent to which a must be raised in onder to obtain \(x\). \
View solution