Chapter 6: Problem 11
Graph each inequality. Do not use a calculator. $$y<3 x^{2}+2$
Short Answer
Expert verified
Sketch the parabola and shade the region below it.
Step by step solution
01
Understand the inequality form
The given inequality is \( y < 3x^2 + 2 \). This has the form \( y < f(x) \), where \( f(x) = 3x^2 + 2 \). It represents a region below the parabola defined by \( y = 3x^2 + 2 \).
02
Sketch the parabola
First, consider the equality \( y = 3x^2 + 2 \) which is a parabola opening upwards. Identify the vertex and plot a few points to outline its shape. The vertex is at the point \((0, 2)\), because substituting \(x = 0\) into \(3x^2 + 2\) gives \(y = 2\). Notice that \(a = 3\) means the parabola is vertically stretched.
03
Identify key points
Choose points symmetric around the vertex, such as \( x = -1, 0, 1 \). Calculate corresponding \( y \) values: \(-1, 3(-1)^2 + 2 = 5\); \( 0, 2 \); \( 1, 3(1)^2 + 2 = 5\). These give points \((-1, 5)\), \((0, 2)\), \((1, 5)\).
04
Draw the boundary parabola
Draw a dashed curve through the points \((-1, 5)\), \((0, 2)\), and \((1, 5)\) to represent the boundary \( y = 3x^2 + 2 \). Use a dashed line to indicate that the inequality is \(<\), not \(\leq\).
05
Shade the solution region
Since the inequality is \( y < 3x^2 + 2 \), shade the region below the parabola. This region represents all the \((x, y)\) points where \( y \) is less than \( 3x^2 + 2 \). Confirm the region by testing a point such as \( (0, 0) \) to ensure it satisfies the inequality.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Inequalities
Quadratic inequalities involve expressions like \( y < ax^2 + bx + c \). These types of inequalities define a region within a coordinate plane related to a quadratic function. The quadratic function forms a parabola, a curve with a distinctive U-shape or upside-down U-shape.
Understanding the graph of this equation helps determine the solution set of a quadratic inequality. In our exercise, the inequality \( y < 3x^2 + 2 \) indicates that we are interested in the region below the parabola defined by \( y = 3x^2 + 2 \).
Understanding the graph of this equation helps determine the solution set of a quadratic inequality. In our exercise, the inequality \( y < 3x^2 + 2 \) indicates that we are interested in the region below the parabola defined by \( y = 3x^2 + 2 \).
- The inequality symbol \(<\) tells us the solution includes all points \((x, y)\) below the curve, not on it.
- A parabola opens upwards when the quadratic coefficient \(a\) is positive and downwards when negative.
Parabola Graphing
Graphing a parabola involves plotting key features of the quadratic function \( y = ax^2 + bx + c \).
The parabola's vertex and symmetry play key roles in accurately sketching its graph. For the example \( y = 3x^2 + 2 \), we can note the parabola opens upwards because the coefficient \( a = 3 \) is positive.
To start graphing, follow these steps:
The parabola's vertex and symmetry play key roles in accurately sketching its graph. For the example \( y = 3x^2 + 2 \), we can note the parabola opens upwards because the coefficient \( a = 3 \) is positive.
To start graphing, follow these steps:
- Identify the vertex of the parabola. Here, the vertex is \((0, 2)\).
- Select additional points, both to the left and right of the vertex, ensuring symmetry. Points like \((-1, 5)\) and \((1, 5)\) work well.
- Draw the parabola smoothly through these points, with each side evenly extending from the vertex.
Vertex Identification
The vertex is a critical feature of any parabola, often marking its turning point. To determine the vertex, use the quadratic function's standard form \( y = ax^2 + bx + c \).
For a basic quadratic equation in this form, the vertex for \( y = ax^2 + c \) directly indicates it lies at \( x = 0 \), making finding the vertex straightforward.
For the function \( y = 3x^2 + 2 \), the constant term alongside \( x^2 \) makes finding the vertex simple:
For a basic quadratic equation in this form, the vertex for \( y = ax^2 + c \) directly indicates it lies at \( x = 0 \), making finding the vertex straightforward.
For the function \( y = 3x^2 + 2 \), the constant term alongside \( x^2 \) makes finding the vertex simple:
- The vertex \( (h, k) \) where \( x = h \) can be found when \( b = 0 \). In this case, it's \((0, c) = (0, 2)\).
- Thus, the vertex is at the point \((0, 2)\), which is perfectly centered along the vertical line of symmetry \( x = 0 \).