Graphs Of Equations X^2+y^2=2 And Y=2x^2-3

Hey guys! Let's dive into a fascinating problem involving a system of equations and their graphical representations. We've got a system that looks like this:

\begin{array}{l}
x^2+y^2=2 \\
y=2 x^2-3
\end{array}

Our mission is to figure out what these equations represent graphically and understand how they interact. So, let's put on our thinking caps and get started!

Unveiling the Equations: A Circle and a Parabola

To understand the system, we first need to identify the equations. The first equation, x² + y² = 2, is a classic example of a circle. Remember the general form of a circle's equation: (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. In our case, we can rewrite the equation as (x - 0)² + (y - 0)² = (√2)². This tells us that we're dealing with a circle centered at the origin (0, 0) and having a radius of √2. Visualizing this, you can imagine a circle neatly drawn around the center of our coordinate plane, stretching out a little beyond 1 on both the x and y axes, since the square root of 2 is approximately 1.414. This circle is the set of all points that are exactly √2 units away from the origin, forming a continuous loop.

Now, let’s look at the second equation: y = 2x² - 3. This equation, my friends, represents a parabola. Parabolas are U-shaped curves, and their standard form often looks like y = ax² + bx + c. In our case, we have a = 2, b = 0, and c = -3. The 'a' value (which is 2) tells us that the parabola opens upwards (since it's positive) and that it is narrower than the standard parabola y = x². The 'c' value (which is -3) tells us that the vertex, or the lowest point of our U-shaped curve, is located at (0, -3). Think of it as the parabola hanging down a bit on the y-axis before it curves back up. The parabola y = 2x² - 3 is a transformation of the basic parabola y = x². The 2 in front of the x² causes a vertical stretch, making the parabola skinnier, and the -3 shifts the entire parabola down by 3 units. This means the lowest point of the parabola, the vertex, is at (0, -3), significantly lower than the origin.

Visualizing the Intersection: Where Circle Meets Parabola

Now that we know what each equation represents, the next step is to visualize how they might intersect. We have a circle centered at the origin and a parabola opening upwards, with its vertex below the x-axis. The crucial question is: how many times do these two shapes cross each other? This will give us the number of real solutions to the system of equations.

Imagine the circle. It's a smooth, continuous curve looping around the origin. Now picture the parabola. It starts low, curves upwards, and gets wider as it goes. The parabola's vertex is at (0, -3), which is below the bottom of our circle. As the parabola opens upwards, it will intersect the circle at two points. Think of the parabola as first dipping below the circle, then as it widens and rises, it slices through the circle on both the left and right sides. These intersection points are the solutions to our system of equations, the points (x, y) that satisfy both the circle equation and the parabola equation simultaneously.

To find these points, we need to solve the system algebraically. This involves substituting one equation into the other to eliminate one variable and solve for the remaining one. Once we have the values for one variable, we can plug them back into either equation to find the corresponding values for the other variable. These pairs of values are the coordinates of the intersection points, the spots where the circle and the parabola shake hands.

Solving the System: Finding the Points of Intersection

Alright, let's roll up our sleeves and solve this system of equations algebraically. We've got:

x^2 + y^2 = 2
y = 2x^2 - 3

The smartest move here is to use substitution. Since we already have 'y' isolated in the second equation, we can substitute '2x² - 3' for 'y' in the first equation. This will give us an equation with only 'x' as the variable, which we can then solve. It's like taking one piece of the puzzle and fitting it perfectly into the other.

So, let's do the substitution. Replacing 'y' in the first equation, we get:

x^2 + (2x^2 - 3)^2 = 2

Now we've got an equation that looks a bit more complex, but don't worry, we'll break it down step by step. First, we need to expand that squared term (2x² - 3)². Remember that (a - b)² = a² - 2ab + b². Applying this, we get:

(2x^2 - 3)^2 = (2x^2)^2 - 2(2x^2)(3) + 3^2 = 4x^4 - 12x^2 + 9

Now we can substitute this back into our equation:

x^2 + 4x^4 - 12x^2 + 9 = 2

Let's simplify and rearrange this into a standard polynomial form. Combine like terms and subtract 2 from both sides to set the equation to zero:

4x^4 - 11x^2 + 7 = 0

This is a quartic equation (degree 4), but it has a special form. Notice that the exponents are 4 and 2, which suggests we can treat this as a quadratic equation in disguise. We can use a clever substitution to make it clearer. Let's set z = x². Then, our equation becomes:

4z^2 - 11z + 7 = 0

Ah, much better! This is a quadratic equation in 'z', which we can solve using factoring, the quadratic formula, or completing the square. Let's try factoring. We're looking for two numbers that multiply to (4 * 7 = 28) and add up to -11. Those numbers are -4 and -7. So, we can rewrite the middle term:

4z^2 - 4z - 7z + 7 = 0

Now, factor by grouping:

4z(z - 1) - 7(z - 1) = 0
(4z - 7)(z - 1) = 0

This gives us two possible values for 'z':

4z - 7 = 0  =>  z = 7/4
z - 1 = 0   =>  z = 1

But remember, z = x². So, we need to solve for 'x':

For z = 7/4:

x^2 = 7/4
x = ±√(7/4) = ±√7 / 2

For z = 1:

x^2 = 1
x = ±1

So, we have four possible x-values: √7 / 2, -√7 / 2, 1, and -1. Now, we need to find the corresponding y-values. We can use the simpler equation y = 2x² - 3.

For x = √7 / 2:

y = 2(√7 / 2)^2 - 3 = 2(7/4) - 3 = 7/2 - 3 = 1/2

For x = -√7 / 2:

y = 2(-√7 / 2)^2 - 3 = 2(7/4) - 3 = 7/2 - 3 = 1/2

For x = 1:

y = 2(1)^2 - 3 = 2 - 3 = -1

For x = -1:

y = 2(-1)^2 - 3 = 2 - 3 = -1

So, our points of intersection are: (√7 / 2, 1/2), (-√7 / 2, 1/2), (1, -1), and (-1, -1). These are the four points where the circle and the parabola meet!

Describing the System: Putting It All Together

Okay, we've done the hard work. We identified the equations, visualized their graphs, and solved the system algebraically. Now, let's put it all together and describe the system in a clear and concise way.

We started with the system:

x^2 + y^2 = 2
y = 2x^2 - 3

We determined that the first equation represents a circle centered at (0, 0) with a radius of √2. The second equation represents a parabola opening upwards with a vertex at (0, -3).

By solving the system, we found four points of intersection: (√7 / 2, 1/2), (-√7 / 2, 1/2), (1, -1), and (-1, -1). This means that the circle and the parabola intersect at four distinct points on the coordinate plane.

So, to describe the system, we can say that it consists of a circle and a parabola that intersect at four points. The circle is centered at the origin and has a radius of √2, while the parabola opens upwards with its vertex at (0, -3). The solutions to the system are the coordinates of the intersection points we found.

Wrapping Up: A Journey Through Equations and Graphs

Guys, we've journeyed through the world of equations and graphs, and what a trip it has been! We started with a system of equations, identified the shapes they represent, visualized their interaction, and then used algebra to pinpoint their intersection points. We've seen how equations can tell stories about shapes, and how solving systems can reveal the connections between those shapes.

This problem beautifully illustrates the interplay between algebra and geometry. By understanding the equations, we could predict the graphical behavior, and by visualizing the graphs, we could anticipate the algebraic solutions. It's this kind of thinking that makes math so powerful and so cool!

So, next time you see a system of equations, remember the journey we took today. Think about the shapes, visualize the graphs, and dive into the algebra. You might just discover something amazing!