Problem 12 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-20: Sketch the graph of the equation, and label the \(x\) - and \(y\)-intercepts. $$ x=2 y^{2}-4 $$

Step-by-Step Solution

Answer

The graph is a sideways parabola opening to the right with x-intercept \((-4, 0)\) and y-intercepts \((0, \sqrt{2})\) and \((0, -\sqrt{2})\).

Step 1: Rewrite the Equation

Start by rewriting the equation for clarity. The given equation is \( x = 2y^2 - 4 \).

Step 2: Find the x-intercepts

To find the x-intercepts, set \( y = 0 \) in the equation and solve for \( x \): \[x = 2(0)^2 - 4 \rightarrow x = -4.\] Thus, the x-intercept is \((-4, 0)\).

Step 3: Find the y-intercepts

To find the y-intercepts, set \( x = 0 \) in the equation and solve for \( y \): \[0 = 2y^2 - 4 \rightarrow 2y^2 = 4 \rightarrow y^2 = 2 \rightarrow y = \pm \sqrt{2}.\]There are two y-intercepts: \((0, \sqrt{2})\) and \((0, -\sqrt{2})\).

Step 4: Sketch the Graph

The equation \( x = 2y^2 - 4 \) is a sideways parabola opening to the right. The vertex is at \((-4, 0)\). The x-intercept is at \((-4, 0)\), and the y-intercepts are at \((0, \sqrt{2})\) and \((0, -\sqrt{2})\). Draw the graph by sketching a parabola with its vertex at \(-4\) on the x-axis.

Key Concepts

Understanding X-Intercepts

An x-intercept is where a graph crosses the x-axis. At this point, the value of y is always zero. In our exercise, the equation is given as \(x = 2y^2 - 4\). To find the x-intercept, replace the y in the equation with zero and solve for x. Doing the math, you get:

• \(x = 2(0)^2 - 4 = -4\)

Thus, our x-intercept is at the point \((-4, 0)\).
Remember, there can be more than one x-intercept, but in this specific parabola equation, there's only one. Being able to find x-intercepts is crucial for graph-sketching, offering a clear starting point for where our parabola touches the x-axis.

Exploring Y-Intercepts

The y-intercepts of a graph occur where the curve crosses the y-axis. At these points, the x-value is always zero. For our sideways parabola, \(x = 2y^2 - 4\), we set \(x = 0\) and solve for y.

• \(0 = 2y^2 - 4\)
• Rearranging gives: \(2y^2 = 4\)
• Simplifying further, \(y^2 = 2\)
• Thus, \(y = \pm\sqrt{2}\)

These calculations reveal two y-intercepts: \((0, \sqrt{2})\) and \((0, -\sqrt{2})\).
Y-intercepts are equally important as they show critical points where the graph crosses the y-axis. In this equation's graph, the parabola spans these two points vertically.

Sideways Parabolas

When we think of a parabola, we often picture a U-shaped curve that opens upward or downward. A sideways parabola can open to the left or the right. Our equation, \(x = 2y^2 - 4\), describes a sideways parabola opening towards the right.
Sideways parabolas have the standard form \(x = ay^2 + by + c\). The vertex, or turning point, of this parabola is at \((-4, 0)\).
When graphing sideways parabolas, it helps to:

• Identify the vertex, which acts as the anchor of the graph.
• Locate important intercepts, both x and y, to guide the sketch.
• Keep in mind the direction the parabola opens; in this case, rightwards.

Understanding the characteristics of sideways parabolas is essential for sketching and analyzing their behaviors. It provides a more comprehensive insight into equations pointing to such graph types.