Problem 8 - Algebra and Trigonometry with Analytic Geometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Equation

The equation given is \( y = -x^2 + 2 \). This is a quadratic equation, which represents a parabolic curve on a graph. The negative sign in front of \( x^2 \) indicates that the parabola opens downwards.

Step 2: Find the Y-Intercept

To find the y-intercept, set \( x = 0 \) in the equation: \( y = -(0)^2 + 2 = 2 \). Therefore, the y-intercept is at the point \( (0, 2) \).

Step 3: Find the X-Intercepts

To find the x-intercepts, set \( y = 0 \) and solve for \( x \): \( 0 = -x^2 + 2 \). Rearrange to get \( x^2 = 2 \). Taking the square root gives \( x = \pm \sqrt{2} \), so the x-intercepts are at the points \( (\sqrt{2}, 0) \) and \( (-\sqrt{2}, 0) \).

Step 4: Sketch the Graph

First, plot the y-intercept \( (0, 2) \) and x-intercepts \( (\sqrt{2}, 0) \) and \( (-\sqrt{2}, 0) \). Then draw a downward-opening parabola passing through these points. The vertex of the parabola, given the standard form of the quadratic equation \( y = -x^2 + 2 \), is at the point \( (0, 2) \), which is also the y-intercept. The symmetry axis is the y-axis because the equation has no linear term in \( x \).

Key Concepts

Understanding X-intercepts

The term **x-intercepts** refers to the points where a graph crosses the x-axis. For the quadratic function given, these are the solutions for the equation when the output value \( y = 0 \). When the equation \( y = -x^2 + 2 \) is set to zero, it becomes \( -x^2 + 2 = 0 \). Solving for \( x \), we rearrange the equation to \( x^2 = 2 \), and then take the square root of both sides.

• \( x = \sqrt{2} \)
• \( x = -\sqrt{2} \)

This gives us two x-intercepts at the points \( (\sqrt{2},0) \) and \( (-\sqrt{2},0) \). These x-intercepts are symmetric about the y-axis, a common feature for quadratic equations. Understanding x-intercepts is crucial when graphing because they help us visualize where the curve of a parabola touches or crosses the horizontal axis.

Exploring Y-intercepts

The **y-intercept** of a graph occurs where it crosses the y-axis. This happens when \( x = 0 \). For the equation \( y = -x^2 + 2 \), substituting \( x = 0 \) results in:

• \( y = -(0)^2 + 2 = 2 \)

Hence, the y-intercept is the point \( (0, 2) \). This particular value represents the maximum point of our downward-opening parabola, because the equation is negative and the vertex is at the y-intercept. The y-intercept is vital for sketching the graph since it indicates the initial starting point of the curve as it moves along the axis.

Graphing Parabolas

A **parabola** is a symmetric curve that is either U-shaped or inverted depending on the sign of its leading term. In our quadratic equation \( y = -x^2 + 2 \), the negative sign indicates the parabola opens downward. Key attributes when graphing a parabola include:

• The vertex: the highest or lowest point (for our graph, it's the highest), located at \( (0, 2) \).
• The axis of symmetry: a vertical line passing through the vertex, here it's the y-axis \( x = 0 \).
• Intercepts: x-intercepts at \( (\sqrt{2}, 0) \), \( (-\sqrt{2}, 0) \) and the y-intercept at (0, 2).

When we draw the parabola on a graph, it should be symmetric with respect to the y-axis due to its nature. Starting from the vertex, plot the intercepts and draw a smooth curve that passes through them, ensuring the parabola curves downwards, opening outwards.