Problem 65 - A Graphical Approach to Precalculus with Limits

Question

Solve each system graphically. Check your solutions. Do not use a calculator. $$\begin{aligned}&x^{2}-y=0\\\&x^{2}+y^{2}=2\end{aligned}$$

Step-by-Step Solution

Answer

The solutions are \((1, 1)\) and \((-1, 1)\).

Step 1: Understand the equations

We have two equations, one is quadratic and another is a circle equation: \(x^2 - y = 0\) and \(x^2 + y^2 = 2\). We'll graph these equations one by one.

Step 2: Graph the first equation

Starting with the first equation, \(x^2 = y\). This is a parabola that opens upwards and its vertex is at the origin \((0, 0)\).

Step 3: Graph the second equation

The second equation \(x^2 + y^2 = 2\) is a circle centered at the origin with a radius of \(\sqrt{2}\).

Step 4: Identify intersection points

We need to find the points where the graphs of the parabola and the circle intersect. These are solutions to the system.

Step 5: Solve algebraically to check

Substitute \(y = x^2\) from the first equation into the second equation: \(x^2 + (x^2)^2 = 2\), simplifying to \(x^2 + x^4 = 2\). This gives a polynomial equation \(x^4 + x^2 - 2 = 0\).

Step 6: Factor and solve the polynomial

Factor the equation \(x^4 + x^2 - 2 = 0\) as \((x^2 - 1)(x^2 + 2) = 0\). Solve \(x^2 - 1 = 0\) to get \(x = \pm 1\) and note that \(x^2 + 2 = 0\) has no real solutions.

Step 7: Solve for y-coordinates

For \(x = 1\) and \(x = -1\), substitute back into \(y = x^2\): \(y = 1\). The intersection points are \((1, 1)\) and \((-1, 1)\).

Step 8: Confirm on graph

Checking on the graph, both points \((1, 1)\) and \((-1, 1)\) lie on the circle and the parabola, confirming that these are solutions.

Key Concepts

Quadratic Equation

The quadratic equation is an essential concept in algebra and can take many forms. In this exercise, the quadratic equation is given as \(x^2 - y = 0\), which can be rewritten as \(y = x^2\). This representation of a quadratic equation forms a parabola when graphed.

• The graph of \(y = x^2\) consists of a symmetrical curve that opens upwards.
• The vertex of this parabola is at the origin \(0, 0\).
• To graph it, simply plot points by calculating \(y\) for various \(x\) values.

Understanding quadratic equations is crucial as they frequently appear in various real-world applications including physics and engineering.

Circle Equation

Circle equations in algebra describe all the points that are equidistant from a center point. In this problem, we have the circle equation \(x^2 + y^2 = 2\).

• This equation represents a circle centered at the origin \(0, 0\).
• The radius of this circle is the square root of 2, \(\sqrt{2}\), because \(x^2 + y^2 = r^2\) gives us \(r = \sqrt{2}\).

To graph a circle, it is crucial to understand its center and radius, as these determine its position and size on the coordinate plane.Visualizing these equations helps greatly in understanding the spatial relationships in geometric contexts.

Intersection Points

Intersection points are crucial when solving systems of equations graphically, as they represent solutions satisfying both equations. Here, we seek where the parabola \(y = x^2\) intersects with the circle \(x^2 + y^2 = 2\).

• Graphically, intersection points appear where the curves of both equations overlap.
• For this exercise, the intersection points are computed algebraically after plotting the graphs.

Identifying and calculating intersection points is essential for solving many real-life problems where different conditions must be satisfied simultaneously, such as in engineering designs and optimization issues.

Polynomial Factoring

Factoring polynomials is a valuable technique in algebra used to simplify expressions and solve equations. In the provided solution, after substituting \(y = x^2\) into the circle equation, we get a polynomial equation: \(x^4 + x^2 - 2 = 0\).

• Factoring breaks down this polynomial into simpler components: \((x^2 - 1)(x^2 + 2) = 0\).
• These factors make it easier to solve for \(x\) values, where \(x^2 - 1 = 0\) yields real solutions, and \(x^2 + 2 = 0\) does not.

Polynomial factoring is powerful, allowing us to solve higher-degree equations easily and is fundamental in both basic and advanced mathematical problem-solving.

Graphing Parabolas

Graphing parabolas helps with understanding the behavior and characteristics of quadratic equations. When graphing the equation \(y = x^2\):

• Plot several points by choosing values for \(x\) and calculating \(y\).
• The graph is symmetric around the y-axis, making it easy to plot.
• You'll notice it has a U-shape and opens upwards, known for its vertex at the origin in this case.

Graphing is crucial as it allows students to visualize equations, aiding in better comprehension of how algebraic concepts translate into geometric forms. This is especially useful when solving complex systems graphically where graphical intuition plays a key role.