Problem 30 - Precalculus : Building Concepts and Connections

Question

Use transformations to graph the quadratic function and find the vertex of the associated parabola. $$g(s)=-(s-2)^{2}+2$$

Step-by-Step Solution

Answer

The vertex of the parabola is at (2, 2) and the graph is a parabola opening downwards.

Step 1: Translate the Function

The quadratic equation given is already in the form \(f(x) = a(x-h)^2 + k\). Here, 'a' is -1, 'h' is 2 and 'k' is 2. This tells us that the basic quadratic function \(f(x) = x^2\) has been horizontally shifted right by 2 units, vertically shifted up by 2 units and then vertically reflected because 'a' is negative.

Step 2: Identify the Vertex

The vertex of the parabola is given by the (h, k) values of -1 and 2 respectively which in this case is (2, 2).

Step 3: Graph the Function

Begin by plotting the vertex (2, 2) on the graph. Since the quadratic is negatively oriented (opens down), draw the sides of the parabola so they open downwards from the vertex point.

Key Concepts

Parabola Transformations

Quadratic functions often come in the form of transformations. Understanding these transformations helps immensely in graphing them accurately. The function we have here is written as \(g(s)=-(s-2)^{2}+2\), which represents a lot of information about its graph.First, consider the form \(f(x) = a(x-h)^2 + k\). Here:

• \(a\) represents vertical stretching or compressing, and when negative, it indicates a reflection over the x-axis.
• \(h\) signifies the horizontal shift from the origin.
• \(k\) indicates the vertical shift.

For \(g(s)=-(s-2)^{2}+2\):

• The negative \(-1\) indicates the parabola is flipped upside down.
• The \(s-2\) shows the graph shifts right by 2 units.
• The \(+2\) indicates a vertical shift upwards by 2 units.

Transformations like these allow you to manipulate the basic shape of \(y=x^2\) to fit various applications.

Graphing Parabolas

Graphing a parabola involves a few key steps that help make the process simpler. Start by identifying essential transformations, then move on to plotting significant points like the vertex. Here's how you can graph \(g(s)=-(s-2)^2+2\):

• Begin with the vertex, \( (2, 2) \), found from the transformation formula.
• Next, consider the direction: since \(a=-1\) is negative, the parabola opens downwards.
• Given the vertex as the starting point, plot symmetric points on either side. This helps visualize the curve efficiently.

Make sure to sketch the curve smoothly and ensure it conforms to the typical "U" shape known for parabolas, but inverted due to the negative sign. If done correctly, you'll have an accurate representation of the function on your graph.

Vertex of a Parabola

The vertex of a parabola is a crucial point that serves as either a peak or a dip on the graph. It is defined by the coordinates \( (h, k) \) in the function \(f(x) = a(x-h)^2 + k\). For our function, \(g(s)=-(s-2)^2+2\), the vertex is clearly given by \( (2, 2) \).The position of the vertex implies:

• The "h" value (2) is where the parabola is horizontally located on the s-axis.
• The "k" value (2) indicates the vertical placement on the same plane.

Graphically, this point stands as the highest point in a downward-opening parabola. Importantly, this vertex is not only a positional marker but also impacts the entire graph's shape and direction. When you have the vertex, you possess a key piece in accurately sketching the parabola.