Problem 10
Question
Sketch the graph of the function defined by the given expression. $$ x^{2}-1 $$
Step-by-Step Solution
Verified Answer
The graph is a parabola opening upwards with vertex (0, -1), passing through (1, 0) and (-1, 0).
1Step 1: 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.
2Step 2: 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.
3Step 3: 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 \( x = 0 \).
4Step 4: 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 \( (0, -1) \).
5Step 5: 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 \( (-1, 0) \).
6Step 6: 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.
Key Concepts
ParabolaVertexAxis of SymmetryY-Intercept
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.
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 \( (0, c) \).
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.
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 \((0, -1)\).
Other exercises in this chapter
Problem 10
Calculate the given expression without using a calculator. \(\cos (2 \pi / 3) \csc (2 \pi / 3)\)
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Write the point-slope equation of the line determined by the two given points. (12,1),(-4,-4)
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The Cartesian equation of a circle is given. Sketch the circle and specify its center and radius. \((x-3)^{2}+y^{2}+y=1\)
View solution Problem 10
Use long division to convert the rational fraction to a (possibly nonterminating) decimal with a repeating block. Identify the repeating block. \(2 / 7\)
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