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In Exercises \(35-44,\) graph the quadratic function. $$f(x)=\frac{1}{2} x^{2}-\frac{1}{2}$$

Short Answer

Expert verified
The graph is an upward-opening parabola with vertex \((0, -\frac{1}{2})\) and axis \(x=0\).

Step by step solution

01

Identify the basic form of the quadratic function

The given function is \( f(x) = \frac{1}{2}x^{2} - \frac{1}{2} \). Recognize that this is a quadratic function in the standard form \( f(x) = ax^2 + bx + c \) where \( a = \frac{1}{2} \), \( b = 0 \), and \( c = -\frac{1}{2} \). Because \( a > 0 \), the parabola opens upwards.
02

Determine the vertex of the parabola

For the function \( f(x) = ax^2 + bx + c \), the vertex \( (h, k) \) can be found using \( h = -\frac{b}{2a} \). Since \( b = 0 \), \( h = 0 \). Substitute \( x = 0 \) into the function to find \( k \): \( k = \frac{1}{2}(0)^2 - \frac{1}{2} = -\frac{1}{2} \). Thus, the vertex is at \((0, -\frac{1}{2})\).
03

Find the axis of symmetry

The axis of symmetry for a parabola in the form \( ax^2 + bx + c \) is given by the line \( x = h \). From Step 2, we have \( h = 0 \), so the axis of symmetry is the line \( x = 0 \).
04

Calculate additional points

To better graph the function, calculate additional points. Substitute \( x = -2, -1, 1, 2 \) into \( f(x) \). For \( x = 1 \), \( f(1) = \frac{1}{2}(1)^2 - \frac{1}{2} = 0 \). Similarly, calculate: \( f(-1) = 0 \), \( f(2) = 1.5 \), \( f(-2) = 1.5 \). This gives points \((-2, 1.5)\), \((-1, 0)\), \((1, 0)\), \((2, 1.5)\).
05

Sketch the graph

Plot the vertex \((0, -\frac{1}{2})\) and the points found in Step 4 on the coordinate plane. Draw the parabola, ensuring it passes through all identified points and is symmetric about the axis \( x = 0 \). The graph is a "U" shaped curve that opens upwards, with the vertex being the lowest point.

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Key Concepts

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

Vertex of a Quadratic Function
The vertex of a quadratic function is a key feature that helps us understand the graph's shape and position. For a function in the form \( f(x) = ax^2 + bx + c \), the vertex can be found using the formula
  • \( h = -\frac{b}{2a} \)
  • \( k = f(h) \)
The values \( h \) and \( k \) together give us the vertex coordinates \((h, k)\). For the function given, \( f(x) = \frac{1}{2}x^2 - \frac{1}{2} \), the vertex is easily calculated because \( b = 0 \). This means \( h = 0 \).
To find \( k \), substitute \( h = 0 \) back into the function: \( k = \frac{1}{2}(0)^2 - \frac{1}{2} = -\frac{1}{2} \). Thus, the vertex is located at \((0, -\frac{1}{2})\), serving as the parabola's lowest point when \( a > 0 \).
Axis of Symmetry
The axis of symmetry is an imaginary line that divides a parabola into two mirror-image halves. This line runs vertically through the vertex of the quadratic function. For the quadratic equation \( ax^2 + bx + c \), the axis of symmetry can be determined using the equation:
  • \( x = -\frac{b}{2a} \)
For our example, \( b = 0 \) simplifies the calculation immediately to \( x = 0 \).
Consequently, the axis of symmetry is the vertical line \( x = 0 \). This line acts like a mirror, ensuring that for every point on one side of the parabola, there is a corresponding point on the opposite side at equal distance from the axis.
It's essential for sketching a precise graph, as it helps ensure that the parabola is balanced on both sides.
Parabola
A parabola is the shape of the graph of a quadratic function. It can open upwards or downwards, depending on the sign of the coefficient \( a \). In our function, \( f(x) = \frac{1}{2}x^2 - \frac{1}{2} \), \( a = \frac{1}{2} \), which is greater than zero.
This means the parabola opens upwards, creating a "U" shape. The key features of a parabola include:
  • The vertex, which represents either the minimum or maximum point—here, it’s the lowest point.
  • The axis of symmetry, a line that ensures the parabola is symmetrical.
  • The direction of opening determined by the sign of \( a \).
The width of the parabola is also affected by \( a \)—a larger \( |a| \) makes a steeper parabola, while a smaller \( |a| \) results in a wider shape. To sketch this specific parabola, identify key points by substituting values into the function to see how it changes. The function should appear as a graceful "U" with its lowest point at the vertex.

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