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The vertex form of a parabola is a very informative way to express the equation of a parabola, especially when you know its vertex. The vertex form is written as:\[(y - k)^2 = 4p(x - h)\]where

• \( (h, k) \) is the vertex of the parabola, which tells us where the parabola "peaks" or "dips."
• \( p \) is a parameter that indicates the distance from the vertex to the focus of the parabola along the axis of symmetry.

Knowing the vertex form, you can easily graph the parabola or understand its key properties. For example, in the given problem, the vertex is at \((-3, 5)\), and that location is directly used in the equation. This form makes it straightforward to visualize or sketch the parabola on a graph.

The standard form of a parabola helps in writing a precise quadratic equation in a more algebraic expression. For a parabola with a horizontal orientation, as in this scenario, the standard formula is:\[(y - k)^2 = 4p(x - h)\]where

• \( (h, k) \) is the vertex.
• \( p \) is a constant dictating the parabola's "width", with its value affecting the parabola's openness.

This form makes it possible to quickly substitute known values, such as a point the parabola passes through. In the exercise, the equation started with this form, and by introducing the point \((5, 9)\), you could solve for \( p \).

The axis of symmetry of a parabola is an imaginary line that divides it into two mirror-image halves. For parabolas oriented horizontally, this axis is typically parallel to the \( x \)-axis. In the exercise, the axis was described as being parallel to the \( x \)-axis, suggesting a horizontal parabola. This line plays a crucial role because it passes through the vertex, here at \((-3, 5)\), allowing any point on the parabola to have a symmetric counterpart on the opposite side of the axis. Understanding the axis of symmetry is essential for predicting and plotting points on a parabola without solving additional equations.

Solving quadratic equations is a skill that plays into finding specific values within a parabola's equation. In the exercise, you needed to substitute a known point into the standard equation for the parabola to solve for the constant \( p \).Here's the process used:1. Substitute the \( x \) and \( y \) values of the point into the equation.2. Carefully perform algebraic manipulations to isolate \( p \).3. Solve for \( p \) to learn about the parabola's precise characteristics.In our example, substituting the point \((5, 9)\) into the equation \[(9 - 5)^2 = 4p(5 + 3)\]set up the equation \[16 = 32p\].Solving for \( p \) simplifies the problem to, \( p = \frac{1}{2} \), which could then be re-inserted into the standard equation, helping us find the exact equation for the parabola.