Problem 43 - Algebra and Trigonometry

Question

Use the graphical method to find all solutions of the system of equations, correct to two decimal places. \(\left\\{\begin{array}{l}{\frac{x^{2}}{9}+\frac{y^{2}}{18}=1} \\ {y=-x^{2}+6 x-2}\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions are approximately (0.37, 4.84) and (5.10, 5.00).

Step 1: Identify each equation from the system

The first equation is \( \frac{x^2}{9} + \frac{y^2}{18} = 1 \), which is in the form of an ellipse. The second equation is \( y = -x^2 + 6x - 2 \), which represents a downward-opening parabola.

Step 2: Graph the ellipse

Rewrite the equation of the ellipse in a familiar form: \( \frac{x^2}{3^2} + \frac{y^2}{\sqrt{18}^2} = 1 \). Here, the center of the ellipse is at the origin (0,0), the semi-major axis along the y-axis is \( \sqrt{18} \), and the semi-minor axis along the x-axis is 3. Plot this ellipse centered at the origin.

Step 3: Graph the parabola

The parabola \( y = -x^2 + 6x - 2 \) can be graphed by finding its key features. It has a vertex, which can be found using the formula \( x = -\frac{b}{2a} \). In this case, \( x = \frac{6}{2} = 3 \). Substitute 3 back into the equation to find \( y = -3^2 + 6*3 - 2 = 7 \). The vertex is at (3,7). Sketch this parabola passing through the vertex and opening downwards.

Step 4: Identify the intersection points

Look for points where the graphs of the ellipse and the parabola intersect. These intersection points are the solutions to the system. From the graphical plot, estimate the coordinates to two decimal places.

Step 5: Confirm solutions

After estimating graphically, confirm the intersection coordinates algebraically if necessary for precision. Using numerical methods or calculations could help here to ensure accuracy to two decimal places.

Key Concepts

Ellipse Equation

The ellipse equation given in the exercise is \( \frac{x^2}{9} + \frac{y^2}{18} = 1 \). This equation represents an ellipse centered at the origin (0,0).
To better understand, an ellipse is a flattened circle. It has two axes: the major axis (longest diameter) and the minor axis (shortest diameter).
In our equation:

• The denominator under \( x^2 \), which is 9, provides the square of the semi-minor axis, making it 3.
• The denominator under \( y^2 \), which is 18, provides the square of the semi-major axis, making it \( \sqrt{18} \), or approximately 4.24.

So, the ellipse is elongated more along the y-axis. Graphing this ellipse helps us visualize its span and how it interacts with other graphs, like a parabola.

Parabola Equation

In this task, the parabola equation is \( y = -x^2 + 6x - 2 \). A parabola is a U-shaped curve that can open upwards or downwards.
This specific equation represents a downward-opening parabola. We know it opens downwards because the coefficient in front of \( x^2 \) is negative.
To sketch this parabola, finding its vertex is essential. The vertex formula \( x = -\frac{b}{2a} \) helps here:

• For our equation, \( a = -1 \), \( b = 6 \), giving \( x = \frac{6}{2} = 3 \).
• Substituting back, \( y = -3^2 + 6*3 - 2 = 7 \).

So, the vertex is at point (3,7). This point is the peak of the parabola since it opens downwards. Knowing the vertex and how the parabola opens allows a clear sketch on a graph.

System of Equations Solutions

A system of equations involves finding values that satisfy all equations within the set simultaneously. In our original task, we are finding the solution for:

• The ellipse: \( \frac{x^2}{9} + \frac{y^2}{18} = 1 \)
• The parabola: \( y = -x^2 + 6x - 2 \)

The solutions to this system are specific \((x, y)\) coordinate points where both equations align perfectly on the graph.
Graphical methods are particularly useful here because they allow us to see the visual overlap. By plotting both the ellipse and the parabola, we can pinpoint intersection areas that represent solutions. Each intersection point confirms an \((x, y)\) set that solves both equations.

Intersection Points Graphing

Graphing aids in visualizing and identifying solutions to systems of equations like our ellipse and parabola.
The intersection points of these graphs are indeed the different solution sets. We estimate these points from the graph to two decimal places for precision.
Steps for finding these graphically include:

• Plot the ellipse and parabola precisely using their equations.
• Identify where the two shapes overlap. This overlap represents the solutions.

For accuracy, after graphing, we can confirm the points by substituting them back into each original equation. This ensures they satisfy both the equations as real solutions. If manual graphing is challenging, software tools can provide a more precise plot, making it easier to zoom in on and confirm these intersection points.