Problem 11 - A Graphical Approach to College Algebra

Question

Graph each inequality. Do not use a calculator. $$y<3 x^{2}+2$

Step-by-Step Solution

Answer

Sketch the parabola and shade the region below it.

Step 1: 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 \( y = 3x^2 + 2 \).

Step 2: 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.

Step 3: 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 \((-1, 5)\), \((0, 2)\), \((1, 5)\).

Step 4: 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 \(\leq\).

Step 5: 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.

Key Concepts

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 \( y = 3x^2 + 2 \).

• 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.

To solve the inequality, you primarily graph the related quadratic equation to visualize the solution region. Once the parabola is sketched, shading the area of interest depicts the solution easily. This graphing process provides a visual solution that algebraically satisfying these inequalities would confirm.

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:

• 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.

Always use dashed lines for the boundary curve on \( y < ax^2 + bx + c \). This indicates that edge values are not included in the solution set since \(<\) does not include equality. Once your parabola is drawn, shading underneath it shows the inequality's solution region where \( y \) values are smaller than the quadratic expression.

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:

• 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 \( x = 0 \).

By identifying the vertex, one can easily sketch and position the graph, ensuring accuracy in depicting the parabola and any related inequality solutions.