Chapter 1: Problem 4
Sketch the graph of the function. $$ f(x)=x^{2}+2 $$
Short Answer
Expert verified
The graph is a vertical parabola opening upwards with vertex at \((0, 2)\).
Step by step solution
01
Identify the Function Type
The given function is \( f(x) = x^2 + 2 \), which is a quadratic function. This is characterized by its parabolic shape, with the general form \( ax^2 + bx + c \). Here, \( a = 1 \), \( b = 0 \), and \( c = 2 \). The parabola opens upwards because \( a > 0 \).
02
Determine the Vertex of the Parabola
The vertex form of a quadratic function is \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex. For the function \( f(x) = x^2 + 2 \), \( h = 0 \) and \( k = 2 \) because it can be rewritten as \( f(x) = (x-0)^2 + 2 \). Thus, the vertex is \((0, 2)\).
03
Identify the Axis of Symmetry
For any quadratic function, the axis of symmetry can be found using \( x = h \). Since \( h = 0 \), the axis of symmetry for \( f(x) = x^2 + 2 \) is the vertical line \( x = 0 \).
04
Calculate Additional Points
To sketch the parabola, calculate additional points on either side of the vertex. For example, substitute \( x = 1 \) and \( x = -1 \) into \( f(x) \) to get \( f(1) = 1^2 + 2 = 3 \) and \( f(-1) = (-1)^2 + 2 = 3 \). These points \((1, 3)\) and \((-1, 3)\) help define the shape.
05
Sketch the Graph
Plot the vertex \((0, 2)\) and the points \((1, 3)\) and \((-1, 3)\) on the coordinate plane. Draw a smooth curve through these points, forming a parabola that opens upward, with the vertex as the minimum point and the axis of symmetry along \( x = 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex of a Parabola
The vertex of a parabola is a fundamental aspect when dealing with quadratic functions. It is the point where the parabola changes direction and can be either a maximum or minimum point based on the orientation of the parabola. Quadratic functions are logically presented in the vertex form as \( f(x) = a(x-h)^2 + k \), where \((h, k)\) directly signals the vertex.
For the function \( f(x) = x^2 + 2 \), we can express it as \( f(x) = (x-0)^2 + 2 \). This makes it clear that the vertex \( (h, k) \) is positioned at \( (0, 2) \). It signifies
For the function \( f(x) = x^2 + 2 \), we can express it as \( f(x) = (x-0)^2 + 2 \). This makes it clear that the vertex \( (h, k) \) is positioned at \( (0, 2) \). It signifies
- the highest or lowest point on the graph,
- the minimum point for upward-facing parabolas like this one,
- the starting point for graphing.
Axis of Symmetry
The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. For quadratics in vertex form, the axis of symmetry is always expressed by \( x = h \).
In the function \( f(x) = x^2 + 2 \), since \( h = 0 \), the axis of symmetry is the line \( x = 0 \). This means:
In the function \( f(x) = x^2 + 2 \), since \( h = 0 \), the axis of symmetry is the line \( x = 0 \). This means:
- the parabola is symmetric with respect to the y-axis,
- each point on one side has a matching point on the other side with equal distance to the axis,
- it is a crucial feature for graph plotting and analyzing symmetry.
Additional Points on Graph
While the vertex is a key starting point, additional points around it help to sketch a more accurate shape of the parabola. By selecting x-values around the vertex and finding their corresponding y-values, we can outline the curvature properly.
For our specific function \( f(x) = x^2 + 2 \), let's look at points like \( x = 1 \) and \( x = -1 \):
For our specific function \( f(x) = x^2 + 2 \), let's look at points like \( x = 1 \) and \( x = -1 \):
- \( f(1) = 1^2 + 2 = 3 \), giving the point \( (1, 3) \),
- \( f(-1) = (-1)^2 + 2 = 3 \), giving the point \( (-1, 3) \).