Chapter 4: Problem 8
Graph the quadratic function. Specify the vertex, axis of symmetry, maximum or minimum value, and intercepts. $$y=2(x+2)^{2}+4$$
Short Answer
Expert verified
Vertex: (-2, 4), axis: x=-2, minimum value: 4, y-intercept: (0, 12). No x-intercepts.
Step by step solution
01
Identify the standard form of a quadratic
The given quadratic equation is in the form of \( y = a(x-h)^2 + k \), where \( a = 2 \), \( h = -2 \), and \( k = 4 \). This is a vertex form of a quadratic equation, where \((h, k)\) represents the vertex.
02
Determine the vertex
From the equation \( y = 2(x+2)^2 + 4 \), the vertex \( (h, k) \) is \( (-2, 4) \). This point is where the parabola changes direction.
03
Find the axis of symmetry
The axis of symmetry for a quadratic equation \( y = a(x-h)^2 + k \) is the vertical line \( x = h \). Therefore, the axis of symmetry is \( x = -2 \).
04
Identify the direction and the maximum/minimum of the parabola
Since \( a = 2 \) which is positive, the parabola opens upwards. Thus, the vertex represents the minimum point of the graph, and the minimum value of the function is \( y = 4 \) which occurs at \( x = -2 \).
05
Calculate the y-intercept
To find the y-intercept, set \( x = 0 \) in the equation: \[y = 2(0+2)^2 + 4 = 2(4) + 4 = 8 + 4 = 12\] The y-intercept is \( (0, 12) \).
06
Find the x-intercepts (if they exist)
Set \( y = 0 \) and solve for \( x \):\[0 = 2(x+2)^2 + 4\]Solving, \[ (x+2)^2 = -2 \]. Since a square cannot be negative, this quadratic function has no real x-intercepts.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex Form
When studying quadratic functions, the vertex form is a specific way of expressing these functions. It is structured as follows:
- The general formula for the vertex form is \( y = a(x-h)^2 + k \), where \( a \) determines the width and direction of the parabola.
- The point \((h, k)\) represents the vertex of the parabola, which is where the curve changes direction.
- Here, \( h = -2 \) and \( k = 4 \), signifying that the vertex of the parabola is the point \((-2, 4)\).
- This informs us of the parabola's lowest point because \( a \) is positive.
Axis of Symmetry
The axis of symmetry in a quadratic function is a vertical line that divides the parabola into two mirror-image halves. It passes directly through the vertex of the parabola:
- For the vertex form, \( y = a(x-h)^2 + k \), the axis of symmetry is always the line \( x = h \).
- The axis of symmetry is \( x = -2 \).
- The parabola is symmetric around \( x = -2 \).
- Every point on the left side of this line corresponds to a point on the right side at an equal distance.
Parabola
A parabola is the U-shaped graph that represents a quadratic function. It holds certain properties:
- It is the shape drawn by plotting a quadratic equation on a graph.
- The vertex, which is the highest or lowest point, depends on whether the parabola opens upwards or downwards.
- Since \( a = 2 \) is positive, the parabola opens upward.
- The vertex \((-2, 4)\) is the minimum point of this parabola.
- Direction (upward or downward)
- Width (influenced by the value of \( a \))
- The position of the vertex and symmetry
Intercepts
Intercepts are points where the parabola intersects the x-axis and y-axis.
Y-intercept
- The y-intercept occurs where the parabola crosses the y-axis. This is found by setting \( x = 0 \) in the equation.
- For \( y = 2(x+2)^2 + 4 \), setting \( x = 0 \) gives \( y = 12 \). Therefore, the y-intercept is \( (0, 12) \).
X-intercepts
- X-intercepts occur where the parabola crosses the x-axis, found by setting \( y = 0 \).
- Solving \( 0 = 2(x+2)^2 + 4 \) results in no real solution, indicating no x-intercepts.