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Chapter 1: Problem 10

Sketch the graph of the function defined by the given expression. $$ x^{2}-1 $$

Short Answer

Expert verified
The graph is a parabola opening upwards with vertex (0, -1), passing through (1, 0) and (-1, 0).

Step by step solution

01

Understand the Function

The function given is \( f(x) = x^2 - 1 \). This is a quadratic function, which can be rewritten in the standard form \( y = ax^2 + bx + c \) where \( a = 1 \), \( b = 0 \), and \( c = -1 \). The graph of a quadratic function is a parabola.
02

Identify the Vertex

Since there is no linear term, the vertex of the parabola is at the point \( (0, c) \) or \( (0, -1) \). This is the lowest point on the parabola since \( a > 0 \), indicating the parabola opens upwards.
03

Find the Axis of Symmetry

The axis of symmetry for a parabola \( y = ax^2 + bx + c \) is the vertical line \( x = -\frac{b}{2a} \). For this function, \( b = 0 \), so the axis of symmetry is \( x = 0 \).
04

Determine the Y-Intercept

The y-intercept occurs where \( x = 0 \). Substituting into the function, \( f(0) = 0^2 - 1 = -1 \), so the y-intercept is \( (0, -1) \).
05

Find Additional Points

Choose values of \( x \) to find other points. For example, if \( x = 1 \), \( f(1) = 1^2 - 1 = 0 \), giving point \( (1, 0) \). Similarly, \( f(-1) = (-1)^2 - 1 = 0 \), providing point \( (-1, 0) \).
06

Sketch the Graph

Plot the vertex, y-intercept, and the points \( (1, 0) \) and \( (-1, 0) \) on a coordinate plane. Draw a smooth curve through these points, ensuring the parabola opens upwards. The vertex at \( (0, -1) \) is the lowest point, and the parabola is symmetric about the y-axis.

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Key Concepts

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

Parabola
Understanding a parabola is key to graphing quadratic functions like the one given: \( f(x) = x^2 - 1 \). In general, the term 'parabola' refers to the U-shaped curve that is the graph of any quadratic function.
  • The shape of a parabola is determined by the sign of the coefficient \( a \) in the function \( ax^2 + bx + c \).
  • If \( a > 0 \), the parabola opens upwards, creating a "smile" shape.
  • If \( a < 0 \), the parabola opens downwards, resembling a "frown" shape.
In this specific case, the coefficient \( a = 1 \) is positive, so the parabola opens upwards, making it one of the simpler cases to understand and sketch.
Vertex
The vertex is a crucial feature of a parabola as it points out the maximum or minimum value of the function, depending on how the parabola opens.
  • The vertex itself is simply a point \((h, k)\) where all sides of the parabola are symmetric.
  • For a standard form function \( ax^2 + bx + c \), if \( b = 0 \), then the vertex will always lie on the y-axis, at \( (0, c) \).
In our exercise with \( f(x) = x^2 - 1 \), the vertex lies at \( (0, -1) \). This is because \( b = 0 \) and \( c = -1 \), meaning the vertex lies on the same line as the y-intercept. Because \( a > 0 \), the vertex is the lowest point on the parabola.
Axis of Symmetry
The axis of symmetry is an imaginary line that divides the parabola into two mirror-image halves. The importance of the axis of symmetry:
  • This vertical line always passes through the vertex of the parabola.
  • For quadratic functions in the form \( ax^2 + bx + c \), it is given by the equation \( x = -\frac{b}{2a} \).
  • This equation balances the parabola, as every point on one side has a corresponding point with the same \( y \)-value on the other side.
In our case, since \( b = 0 \), the axis of symmetry is \( x = 0 \), which matches with the vertex line. As such, the parabola is evenly balanced and appears symmetric about the y-axis.
Y-Intercept
The y-intercept is another fundamental point on the graph of a quadratic function. It is simply where the parabola intersects the y-axis.
  • To find the y-intercept, substitute \( x = 0 \) into the function.
  • For our exercise, setting \( x = 0 \) in \( f(x) = x^2 - 1 \) gives \( f(0) = 0^2 - 1 = -1 \).
  • Thus, the y-intercept is at the point \((0, -1)\).
This point also acts as a handy reference for locating the vertex, as both share the same \( y\)-coordinate in our particular problem. Being aligned on the axis of symmetry also simplifies plotting other points for a clearer graph.

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