Problem 7 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-20: Sketch the graph of the equation, and label the \(x\) - and \(y\)-intercepts. $$ y=2 x^{2}-1 $$

Step-by-Step Solution

Answer

The graph is a parabola opening upwards, with vertex (0, -1), y-intercept (0, -1), and x-intercepts \((\frac{\sqrt{2}}{2}, 0)\) and \((-\frac{\sqrt{2}}{2}, 0)\).

Step 1: Identify the Type of Equation

The equation given is \( y = 2x^2 - 1 \). This is a quadratic equation of the form \( y = ax^2 + bx + c \), where \( a = 2 \), \( b = 0 \), and \( c = -1 \). Quadratic equations form a parabola when graphed.

Step 2: Find the Vertex of the Parabola

The vertex form of a quadratic equation is \( y = a(x-h)^2 + k \). Since \( b = 0 \), the vertex is at \( x = \frac{-b}{2a} = 0 \). Evaluate at \( x = 0 \): \( y = 2(0)^2 - 1 = -1 \). Thus, the vertex is at (0, -1).

Step 3: Determine the y-intercept

To find the \( y \)-intercept, set \( x = 0 \). Substitute into the equation: \( y = 2(0)^2 - 1 = -1 \). Therefore, the \( y \)-intercept is at (0, -1).

Step 4: Determine the x-intercepts

To find the \( x \)-intercepts, set \( y = 0 \) and solve for \( x \):\[ 0 = 2x^2 - 1 \] Add 1 to both sides to get \( 2x^2 = 1 \). Divide by 2 to isolate \( x^2 \): \( x^2 = \frac{1}{2} \). Taking the square root of both sides, \( x = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2} \). Thus, the \( x \)-intercepts are \( (\frac{\sqrt{2}}{2}, 0) \) and \( (-\frac{\sqrt{2}}{2}, 0) \).

Step 5: Sketch the Graph

Draw a parabolic curve opening upwards as \( a > 0 \). Label the vertex at (0, -1) and the intercepts as calculated: \( y \)-intercept at (0, -1), and \( x \)-intercepts at \( (\frac{\sqrt{2}}{2}, 0) \) and \( (-\frac{\sqrt{2}}{2}, 0) \). The graph is symmetrical about the y-axis.

Key Concepts

Parabola

A parabola is the graph of a quadratic equation, such as the one given in the exercise, \( y = 2x^2 - 1 \). Parabolas are

• U-shaped graphs that can open either upwards or downwards.
• Symmetrical around a vertical line called the axis of symmetry.
• The point where the parabola is at its minimum or maximum is called the vertex.

In this exercise, our parabola opens upwards because the coefficient of \( x^2 \) is positive (\( a = 2 \)). The vertex of this parabola is located at (0, -1), indicating where the minimum point of the graph lies. Understanding the direction in which a parabola opens and identifying the vertex can be crucial for sketching its graph accurately. Each side of the parabola mirrors the other along the axis of symmetry, which, in this case, is the y-axis.

x-intercepts

The x-intercepts of a graph are the points where the graph touches or crosses the x-axis.

• To find the x-intercepts, you set \( y = 0 \) in the equation.
• For the equation \( y = 2x^2 - 1 \), solving for \( x \) when the equation equals zero gives you the x-intercepts.
• Setting \( 0 = 2x^2 - 1 \) leads to \( x^2 = \frac{1}{2} \).

Taking the square root of both sides results in \( x = \pm \frac{\sqrt{2}}{2} \). Therefore, the x-intercepts are at the points \( (\frac{\sqrt{2}}{2}, 0) \) and \( (-\frac{\sqrt{2}}{2}, 0) \).This means that the parabola crosses the x-axis at these two symmetrical points. Understanding how to find these intercepts not only helps in graphing the quadratic equation but also aids in analyzing the zeros of the function.

y-intercepts

The y-intercept of a graph is where the graph touches or crosses the y-axis. This happens when \( x = 0 \).

• For any equation, setting \( x = 0 \) will give the y-intercept directly.
• In \( y = 2x^2 - 1 \), substituting \( x = 0 \) results in \( y = -1 \).

Therefore, the y-intercept is at the point (0, -1). This intercept provides insight into how the graph behaves as it crosses the y-axis. In our specific parabola, it verifies our vertex point, as the vertex is at the lowest point of the graph due to its upward opening nature. The y-intercept coinciding with the vertex is a unique scenario that occurs here because the parabola is symmetrical about the y-axis, and the axis of symmetry passes through the origin. Recognizing where a parabola crosses the y-axis is essential for sketching it accurately in a coordinate plane.