91Ó°ÊÓ

The vertex form of a quadratic function is a useful representation because it clearly shows the vertex of the parabola. The general formula for a quadratic function in vertex form is: \[ f(x) = a(x - h)^2 + k \]where

• \( (h, k) \) is the vertex of the parabola
• \( a \) is a coefficient that affects the *width* and *direction* of the parabola

A positive \( a \) means the parabola opens upwards, while a negative \( a \) means it opens downwards. This form makes it easy to identify the maximum or minimum point of the graph. For example, in the quadratic function \( f(x) = x^2 - 2x - 3 \), the vertex form is \( f(x) = (x - 1)^2 - 4 \) and the vertex is \( (1, -4) \). Converting standard form into vertex form often involves completing the square. Always remember, changing a quadratic to vertex form simplifies finding the vertex and analyzing the graph's behavior much easier.

The focal point, or focus, of a parabola is an important concept in quadratic functions. It's a point from which distances to any point on the parabola are equal. We calculate the focal point using the formula: \[ d = \frac{1}{4a} \]where

• \( d \) is the distance from the vertex to the focal point
• \( a \) is the coefficient in the quadratic equation

For example, given the quadratic \( f(x) = (x - 1)^2 - 4 \) with \( a = 1 \), the distance to the focal point is \( \frac{1}{4 \times 1} = 0.25 \). Since our parabola opens upwards, the focal point will be directly above the vertex. If the vertex is at \( (1, -4) \), you add the distance to the \( y \)-coordinate: \[ -4 + 0.25 = -3.75 \]The focal point then is at \( (1, -3.75) \). This information is fundamental when graphing parabolas, as it helps in understanding the openness and precise location of the curve's focus.

The distance formula is often used in various topics in mathematics, especially in coordinate geometry. It is particularly useful in our context of finding the focal point. The distance formula calculates the straight-line distance between two points in a plane and is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]where

• \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of the two points

In our exercise, however, the simpler formula \( d = \frac{1}{4a} \) suffices for vertical distance from the vertex to the focal point in a parabola. For the quadratic \( g(t) = 2(t - 4)^2 - 8 \), we use \( a = 2 \). Thus the distance is \[ \frac{1}{4 \times 2} = 0.125 \]The vertex is \( (4, -8) \), and to find the focal point, we add this distance to the \( y \)-coordinate: \[ -8 + 0.125 = -7.875 \]So, the focal point is \( (4, -7.875) \). Being comfortable with the distance formula and its application in different contexts, such as finding focal points, is crucial for mastering quadratic functions.