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A parabola is a U-shaped curve that you often see in graphs of quadratic functions. The simplest parabola in standard form is given by the equation \(y = x^2\). Every parabola has important features:
* **Vertex**: The point where the parabola changes direction. For \(y = x^2\), the vertex is at the origin (0, 0).
* **Axis of symmetry**: The vertical line that divides the parabola into two mirror images. For \(y = x^2\), this line is the y-axis.
* **Direction**: Parabolas can open upward (like \(y = x^2\)), or downward (like \(y = -x^2\)). In our problem, both parabolas open upward.

Parabolas can look different based on their equations. For example, \(y = (x - 3)^2\) represents a parabola that has been shifted 3 units to the right from the origin.

Quadratic equations are polynomial equations of degree 2 in the form \(ax^2 + bx + c = 0\). Here's why they are important:
* **Graph**: Their graphs are always parabolas.
* **Solving**: You can solve these equations to find where their parabolas intersect the x-axis or other curves. For example, to find their roots (solutions for \(x\)), you can use methods like factoring, completing the square, or the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In the given exercise, we dealt with the quadratic equations \(y = x^2\) and \(y = (x-3)^2\). By setting their right-hand sides equal (i.e., \(x^2 = (x-3)^2\)), we solved for their intersection points.

Coordinate geometry, also known as analytic geometry, allows us to describe geometric shapes using algebra and coordinate points. Here's a quick guide:
* **Coordinates**: Points are represented by their coordinates \((x, y)\), showing their position on the x-axis and y-axis.
* **Graphs**: Equations like \(y = x^2\) can be graphed on the Cartesian plane, creating shapes like the parabola.
* **Intersections**: To find where two graphs (like parabolas) intersect, you solve the equations simultaneously.

In this exercise, we found the intersection of two parabolas by solving the equations \(x^2 = (x - 3)^2\) to get the x-coordinates, and then substituting back to get the y-coordinates, resulting in the point \((\frac{3}{2}, \frac{9}{4})\).

Solving equations involves finding the values of variables that satisfy the equations. Here are the steps we used in the exercise:
* **Equate**: Set the equations equal to each other, \(x^2 = (x-3)^2\).
* **Expand and simplify**: Expand \((x-3)^2\) to get \(x^2 = x^2 - 6x + 9\) and simplify to \(0 = -6x + 9\).
* **Solve**: Isolate x by solving \(6x = 9\) giving us \(x = \frac{3}{2}\).
* **Find y**: Substitute back into one of the original equations to find y. Plugging \(x = \frac{3}{2}\) into \(y = x^2\) gives \(y = \frac{9}{4}\).

These steps help us find the point of intersection \((\frac{3}{2}, \frac{9}{4})\) where both parabolas meet.