Problem 80 - A Graphical Approach to Precalculus with Limits

Question

Fill in each blank with the appropriate response. The graph of the system,$$y>x^{2}+2$$ $$x^{2}+y^{2}<16\( $$y<7$$ consists of all points \)\frac{\underline{\phantom{xx}}}{\text { (above/below) }}\( the parabola given by \)y=x^{2}+2, \frac{\underline{\phantom{xx}}}{(\text { insideloutside })}\( the circle \)x^{2}+y^{2}=16,\( and \)\frac{\underline{\phantom{xx}}}{\text { (above/below) }}\( the line \)y=7$.

Step-by-Step Solution

Answer

above, inside, below

Step 1: Analyze the Parabola

The inequality given is \( y > x^2 + 2 \). This inequality indicates that we are considering the region above the parabola \( y = x^2 + 2 \), where for any point \((x, y)\), the \( y \)-coordinate must be greater than \( x^2 + 2 \).

Step 2: Analyze the Circle

The inequality \( x^2 + y^2 < 16 \) suggests that the graph includes points inside the circle with radius 4, centered at the origin. This is because the inequality is strictly less than, making the points strictly inside the circle.

Step 3: Analyze the Line

The inequality \( y < 7 \) tells us that the points must be below the line \( y = 7 \). This means for any point \((x, y)\), the \( y \)-coordinate must be less than 7.

Step 4: Combine the Results

Based on the results of Steps 1, 2, and 3, the graph consists of points that are above the parabola \( y = x^2 + 2 \), inside the circle \( x^2 + y^2 = 16 \), and below the line \( y = 7 \).

Key Concepts

Parabola

A parabola is a curve on a plane that is defined by a quadratic equation. In this particular problem, the parabola is represented by the equation \( y = x^2 + 2 \). The inequality \( y > x^2 + 2 \) implies that we need to consider all the points on the coordinate plane that are above this curve.

• Understanding the parabola helps in identifying the region where the y-values exceed \( x^2 + 2 \).
• Graphically, this means shading the area directly above the curve, where for every x, the y-coordinate is greater than \( x^2 + 2 \).

When graphing this, visualize the parabola bending upwards, moving away from the y-axis as it extends. This particular parabola is quite simple, symmetric around the y-axis, and the vertex is at the point (0, 2) since the equation is in the form \( y = x^2 + c \). Any point you choose above this curve will satisfy \( y > x^2 + 2 \). That defines the region of interest for this part of the inequality.

Circle

A circle can be easily identified on a graph by the general equation \( x^2 + y^2 = r^2 \), where \( r \) is the radius. In this instance, we are dealing with \( x^2 + y^2 = 16 \), meaning we have a circle centered on the origin with a radius of 4. The inequality \( x^2 + y^2 < 16 \) specifies the region inside this circle.

• The circle starts at (0,0) and extends to have a radius of 4 units outward in all directions.
• The goal is to find all points inside, not on the circle boundary.

Plotting this graphically, you start at the center (0,0) and draw a perfect circle reaching out to 4 units away, both up and down, left and right. When you see \( x^2 + y^2 < 16 \), visualize shading everything inside this circle. It excludes the circular rim, focusing only on the inner region. This makes it clear why we're only considering points strictly within, not touching the circle itself.

Linear Inequalities

Linear inequalities involve straight lines when graphed. In the exercise, we're looking at \( y < 7 \). This inequality depicts everything below the line \( y = 7 \).

• Graphing \( y = 7 \) gives a horizontal line crossing the y-axis.
• The inequality emphasizes a region where y-values are less than 7.

Consider the line as a boundary of a region extending infinitely downward. Every point below the line has a y-coordinate that satisfies \( y < 7 \). In practical terms, draw a horizontal line at y = 7 and imagine shading everything downwards. This helps in understanding how any point with a lesser y-value fulfills the condition of the inequality. The line acts as a ceiling that limits the vertical span of acceptable points.