Problem 27
Question
Graph each parabola. Give the vertex, axis of symmetry, domain, and range. $$ f(x)=x^{2}+4 x+3 $$
Step-by-Step Solution
Verified Answer
The vertex is \((0, -1)\). The axis of symmetry is \(x = 0\). The domain is \(( -\infty, \infty )\). The range is \([-1, \infty )\).
1Step 1: Identify the vertex
The general form of a parabola is given by \(f(x) = ax^2 + bx + c\). Here, we have \(a = 1\), \(b = 0\), and \(c = -1\). For a parabola in this form, the vertex can be found using the formula \(\left( -\frac{b}{2a}, f\left( -\frac{b}{2a} \right) \right)\). Since \(b = 0\), the x-coordinate of the vertex is \(0\). Substituting \(x = 0\) into \(f(x)\), we get the y-coordinate as \(f(0) = 0^2 - 1 = -1\). Therefore, the vertex is \((0, -1)\).
2Step 2: Determine the axis of symmetry
The axis of symmetry for a parabola in the form \(f(x) = ax^2 + bx + c\) is given by \(x = -\frac{b}{2a}\). For this parabola, the axis of symmetry is \(x = 0\).
3Step 3: Find the domain
The domain of a parabola represented by \(f(x) = x^2 - 1\) is all real numbers. In interval notation, this is \(( -\infty, \infty )\).
4Step 4: Find the range
The range of the parabola can be determined by observing its vertex. Since the vertex \((0, -1)\) is the minimum point and the parabola opens upward, the range is \([-1, \infty )\).
5Step 5: Graph the parabola
To graph \(f(x) = x^2 - 1\), start by plotting the vertex \((0, -1)\). The axis of symmetry is \(x = 0\). For additional points, evaluate \(f(x)\) at several x-values. For example: \(f(1) = 1^2 - 1 = 0\) and \(f(-1) = (-1)^2 - 1 = 0\). Plot these additional points and draw a smooth curve through them, opening upwards.
Key Concepts
Vertex of a ParabolaAxis of SymmetryDomain and Range
Vertex of a Parabola
Understanding the vertex of a parabola is crucial when graphing. The vertex represents the highest or lowest point on the graph, depending on whether the parabola opens downwards or upwards. For the equation provided, \( f(x) = x^2 - 1 \), the general form is \( f(x) = ax^2 + bx + c \) where \( a = 1 \), \( b = 0 \), and \( c = -1 \). The vertex formula is given by the coordinates \( \left( -\frac{b}{2a}, f\left( -\frac{b}{2a} \right) \right) \). In this case, \( b = 0 \), making the x-coordinate of the vertex \( 0 \). To find the y-coordinate, substitute \( x = 0 \) back into the equation, yielding \( f(0) = 0^2 - 1 = -1 \). Thus, the vertex is \( (0, -1) \). This specific point will serve as a key reference when plotting the parabola. The vertex in this scenario is a minimum point since the coefficient of \( x^2 \) is positive, indicating that the parabola opens upwards.
Axis of Symmetry
The axis of symmetry is an imaginary vertical line that runs through the vertex, dividing the parabola into two mirror-image halves. For any parabola in the form \( ax^2 + bx + c \), the axis of symmetry can be found using the formula \( x = -\frac{b}{2a} \). In the given function \( f(x) = x^2 - 1 \), with \( b = 0 \) and \( a = 1 \), the axis of symmetry is simply \( x = 0 \). This line helps you to understand how the parabola is mirrored and it ensures that whatever happens on one side of \( x = 0 \) also happens on the other side. When graphing, it is useful to draw this line so you can accurately plot symmetrical points and get a precise graph.
Domain and Range
The domain and range of a function tell you the possible inputs and outputs, respectively. For a parabola given by \( f(x) = ax^2 + bx + c \), the domain is all real numbers, as you can substitute any real number for \( x \) and get a corresponding \( y \)-value. Therefore, the domain of \( f(x) = x^2 - 1 \) is \(( -\infty, \infty )\). The range depends on the vertex and whether the parabola opens upwards or downwards. Since our vertex is at \( (0, -1) \) and the parabola opens upwards (Because \( a = 1 \) is positive), the smallest \( y \)-value is \( -1 \). Therefore, the range is \( [-1, \infty ) \). This means that \( y \) can be any value from \( -1 \) going up to positive infinity. Understanding the domain and range helps you to see the full extent of the graph.
Other exercises in this chapter
Problem 26
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