Problem 99
Question
Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. $$y=2 x^{2}-4 x+5$$
Step-by-Step Solution
Verified Answer
Vertex: (1, 3), Axis: x = 1, Domain: (-∞, ∞), Range: [3, ∞).
1Step 1: Identify the Quadratic Form
The given equation is in the form of a quadratic equation: \( y = ax^2 + bx + c \) where \( a = 2 \), \( b = -4 \), and \( c = 5 \). This form is suitable for determining various properties of the parabola.
2Step 2: Find the Vertex
The vertex of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Calculate \( x = -\frac{-4}{2 \times 2} = 1 \). Substitute \( x = 1 \) back into the equation \( y = 2(1)^2 - 4(1) + 5 \) to find \( y = 3 \). Thus, the vertex is \((1, 3)\).
3Step 3: Determine the Axis of Symmetry
The axis of symmetry for a parabola with a known vertex \((h, k)\) is \( x = h \). Therefore, since the vertex is \((1, 3)\), the axis of symmetry is \( x = 1 \).
4Step 4: Define the Domain
For any quadratic function of the form \( y = ax^2 + bx + c \), the domain is all real numbers. Thus, the domain is \((-fty, fty)\).
5Step 5: Define the Range
Since \( a = 2 \) is positive, the parabola opens upwards. The minimum value of \( y \) is the \( y\)-coordinate of the vertex, which is \( 3 \). Therefore, the range of \( y \) is \([3, fty)\).
6Step 6: Plot the Parabola
To sketch the graph, plot the vertex at \((1, 3)\) on the graph and draw the axis of symmetry along \( x = 1 \). Since \( a = 2 \), which is positive, the parabola opens upward. Mark a few more points on the graph by choosing values for \( x \) (like 0, 2, 3) and calculating their corresponding \( y \) values to ensure the parabolic shape is accurate.
7Step 7: Verify Using a Graphing Calculator
Input the equation \( y = 2x^2 - 4x + 5 \) into a graphing calculator. Check that the vertex is at \((1, 3)\), the axis of symmetry is \( x = 1 \), and the parabola opens upwards as expected with the correct domain and range.
Key Concepts
Quadratic EquationVertexAxis of SymmetryDomain and Range
Quadratic Equation
In mathematics, a quadratic equation is any equation that can be rewritten in the form:
- \( y = ax^2 + bx + c \)
- If \( a > 0 \), the parabola opens upwards.
- If \( a < 0 \), the parabola opens downwards.
Vertex
The vertex of a parabola is a critical point that represents the minimum or maximum of the quadratic function. For a parabola of the form \( y = ax^2 + bx + c \), the vertex can be calculated using the vertex formula:
In our example with \( y = 2x^2 - 4x + 5 \):
- \( x = -\frac{b}{2a} \)
In our example with \( y = 2x^2 - 4x + 5 \):
- Calculate \( x = -\frac{-4}{2 \times 2} = 1 \).
- Find \( y \) by substituting \( x = 1 \) back into the equation: \( y = 2(1)^2 - 4(1) + 5 = 3 \).
Axis of Symmetry
The axis of symmetry of a parabola is an imaginary vertical line that divides the parabola into two symmetrical halves. This line passes through the vertex. For a parabola described by \( y = ax^2 + bx + c \), the axis of symmetry is given by the formula:
- \( x = -\frac{b}{2a} \)
- With a vertex at \((1, 3)\), the axis of symmetry is \( x = 1 \).
Domain and Range
The domain and range describe the set of possible inputs (\( x \) values) and outputs (\( y \) values) for a function, respectively. For any quadratic function,
However, the range of the parabola depends on the direction in which it opens. For the quadratic \( y = 2x^2 - 4x + 5 \):
- The domain is always all real numbers \((-\infty, \infty)\).
However, the range of the parabola depends on the direction in which it opens. For the quadratic \( y = 2x^2 - 4x + 5 \):
- The parabola opens upward, indicating all \( y \) values starting from the vertex \( y \)-coordinate.
- Here, the vertex is \((1, 3)\), so the range is \([3, \infty)\).
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