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Parabolas are fascinating shapes that you'll often encounter in algebra and geometry. A parabola is a curve formed by a quadratic equation like \(y = ax^2 + bx + c\). It can either open upwards or downwards depending on whether the coefficient \(a\) is positive or negative. In this problem, the parabola is defined by the equation \(y = 4 - x^2\), indicating a downward-opening shape since the \(x^2\) term is negative.

Key characteristics of a parabola include its vertex: the highest or lowest point on the curve. For our parabola, the vertex is at point \((0, 4)\).

Moreover, the axis of symmetry is a vertical line that passes through the vertex, which in this case, is the y-axis given by \(x = 0\). Such properties make parabolas easy to work with when determining points or graphing the curve.

The domain of a function tells us all the possible x-values that can be inputted into a function without encountering any problems. Here, we are working with the distance from a point \((x, y)\) on the graph of a parabola to a fixed point \((0, 1)\). The function we obtain is represented as \(d(x) = \sqrt{x^4 - 5x^2 + 9}\).

To ensure this function is well-defined, we must confirm that the expression under the square root (often called the radicand) stays non-negative. Fortunately, in this scenario, the polynomial \(x^4 - 5x^2 + 9\) is always positive since it is a quartic (degree 4) polynomial with a positive leading coefficient, ensuring it doesn't dip below 9. Therefore, the domain of this function is all real numbers, denoted as \((-fty, fty)\).

Understanding the domain is crucial as it helps us know where the function is applicable and valid across the x-axis.

The distance formula is a handy tool used to determine the distance between two points in a 2D space. If you know the coordinates \((x_1, y_1)\) and \((x_2, y_2)\) of these points, you can use the formula:

• \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

In this problem, our goal is to measure the distance between a point on the parabola \((x, 4-x^2)\) and the fixed point \((0, 1)\).

Using the distance formula, you substitute the coordinates into the formula, leading to:

• \[d = \sqrt{(x - 0)^2 + ((4-x^2) - 1)^2} = \sqrt{x^2 + (3-x^2)^2}\]

Simplifying this gives \(d = \sqrt{x^4 - 5x^2 + 9}\).

Being comfortable with the distance formula is essential since it enables you to connect various geometric concepts and calculate distances that are vital in both pure and applied mathematics.