Problem 30 - A Graphical Approach to College Algebra

Question

Write an inequality that satisfies the description. Below the parabola with vertex ( \(0,1\) ) and \(x\) -intercepts \((-2,0)\) and \((2,0)\) \((-1,0)\) and \((1,0)\)

Step-by-Step Solution

Answer

The inequality is \(y < -\frac{1}{4}x^2 + 1\).

Step 1: Identifying the Parabola Equation

The parabola with vertex (0, 1) and given x-intercepts can be expressed in vertex form: \(y = a(x-h)^2 + k\). Here, \((h, k)\) is the vertex. So, \(h = 0\) and \(k = 1\). Substituting these values gives \(y = a(x-0)^2 + 1 = ax^2 + 1\).

Step 2: Using the x-intercepts to Find 'a'

For the parabola to pass through points \((2, 0)\) and \((-2, 0)\), plug \(x = 2\) and \(y = 0\) into the equation: \(0 = a(2)^2 + 1 \0 = 4a + 1 \a = -\frac{1}{4}\). So, the equation becomes \(y = -\frac{1}{4}x^2 + 1\).

Step 3: Writing the Inequality

The inequality should describe the region below the parabola. Therefore, since \(y = -\frac{1}{4}x^2 + 1\) is the equation of the parabola, the inequality describing the region below is given by: \(y < -\frac{1}{4}x^2 + 1\). This inequality represents values of \(y\) that are less than the parabolic curve.

Key Concepts

Vertex Form of a Parabola

To easily describe the shape and position of a parabola, we often use the vertex form of a quadratic equation. The vertex form is expressed as \( y = a(x-h)^2 + k \). Here, \((h, k)\) is the vertex of the parabola, which is the highest or lowest point on the graph depending on the parabola's orientation.
- **Iconic Elements**: - \(a\) determines the width and direction of the parabola. A positive \(a\) means the parabola opens upwards, while a negative \(a\) means it opens downwards. - \(h\) and \(k\) move the parabola along the x and y axes, respectively.
Understanding this form allows us to translate and manipulate parabolic equations easily.

Finding the Vertex

Finding the vertex of a parabola is quite straightforward once you know the equation in vertex form, \( y = a(x-h)^2 + k \). The vertex is simply the point \((h, k)\).
- **Practical Example**: - If you have the equation \( y = a(x-0)^2 + 1 \), the vertex would be at (0, 1).
This knowledge helps in graphing the parabola and understanding its position and orientation on the Cartesian plane.

Solving Inequalities

Solving inequalities in the context of parabolas involves determining the region where a certain condition is met, such as being above or below the parabola's curve.
- **Expressing Inequalities**: - If a problem asks for the region below a parabola like \( y = -\frac{1}{4}x^2 + 1 \), the solution is expressed as \( y < -\frac{1}{4}x^2 + 1 \). - This inequality indicates that we are considering all \( y \) values less than the parabola at any given \( x \).
Unearthing solutions in this context involves pinpointing the precise areas that satisfy the inequality.

X-intercepts of a Parabola

The x-intercepts of a parabola are the points where the parabola crosses the x-axis. These are crucial in determining the shape and position of the parabola. In an equation of the form \( ax^2 + bx + c = 0 \), the x-intercepts can be found by setting \( y = 0 \) and solving for \( x \).
- **Location and Significance**: - For our parabola, the intercepts are \((-2, 0)\) and \((2, 0)\). - They share the same \( y \)-value of 0 since they lie on the x-axis.
Identifying x-intercepts is an essential step in graphing and understanding the parabola, enabling the determination of its slope and direction.