Problem 5 - A Graphical Approach to College Algebra

Question

For each quadratic function defined , (a) write the function in the form \(P(x)=a(x-h)^{2}+k,\) (b) give the vertex of the parabola, and (c) graph the function. Do not use a calculator. $$P(x)=x^{2}-2 x-15$$

Step-by-Step Solution

Answer

(a) \(P(x) = (x-1)^2 - 16\), (b) Vertex: (1, -16), (c) Plot the parabola with an upward opening and vertex at (1, -16).

Step 1: Identify the Quadratic Coefficients

Given the quadratic function \(P(x) = x^2 - 2x - 15\), let's identify the coefficients: Here, \(a = 1\), \(b = -2\), and \(c = -15\).

Step 2: Complete the Square

To write the function in the form \(a(x-h)^2+k\), we need to complete the square. Let us first focus on the quadratic and linear terms \(x^2 - 2x\). We take the coefficient of \(x\), which is \(-2\), divide it by 2 to get \(-1\), and square this result, resulting in \( (-1)^2 = 1 \). Add and subtract this square inside the function:\[ P(x) = (x^2 - 2x + 1) - 1 - 15 \] This simplifies to:\[ P(x) = (x - 1)^2 - 16 \] Thus, the quadratic is in the form \(P(x) = a(x-h)^2+k\) where \(a = 1\), \(h = 1\), and \(k = -16\).

Step 3: Determine the Vertex

The vertex form of the quadratic function is \(P(x) = a(x-h)^2 + k\). Here, \(h = 1\) and \(k = -16\), so the vertex of the parabola is \((h, k) = (1, -16)\).

Step 4: Graph the Quadratic Function

While graphing the function \(P(x) = (x-1)^2 - 16\), note the following:- It represents a parabola that opens upwards because \(a = 1 > 0\).- The vertex of the parabola, from Step 3, is \((1, -16)\). - To sketch the graph, plot the vertex and use the symmetry of the parabola around the line \(x = 1\). Also, identify other points by selecting values of \(x\) around the vertex, such as \(x = 0\) and \(x = 2\), and determine the corresponding \(P(x)\) values.

Key Concepts

Vertex Form

The vertex form of a quadratic function is particularly useful for identifying the properties of a parabola. It's expressed as \( P(x) = a(x-h)^2 + k \). Here's what each term represents:

• \(a\): Determines the direction the parabola opens. If \(a > 0\), it opens upwards; if \(a < 0\), it opens downwards.
• \(h\): The x-coordinate of the vertex, indicating the horizontal shift of the vertex from the origin.
• \(k\): The y-coordinate of the vertex, indicating how far up or down the vertex is from the origin.

Together, \((h, k)\) forms the vertex of the parabola. Identifying a quadratic in vertex form is essential because it immediately tells us the most important point on a parabola: its vertex. This form also makes it easier to graph the function since the vertex provides a starting point. The factored square, \((x-h)^2\), around which the parabola is symmetric, guides the shape of the curve.

Complete the Square

Completing the square is a method used to convert a standard form quadratic equation into vertex form. This process involves transforming a quadratic expression like \( x^2 - 2x \) into a perfect square trinomial. Here is a simple guide:

• Take the linear coefficient \(b\), in this example \(-2\), divide it by 2, and square it: \((-2/2)^2 = 1\).
• Add and subtract this squared number inside the equation: \( x^2 - 2x + 1 - 1 = (x-1)^2 \).

Incorporating these steps changes \( x^2 - 2x \) to a perfect square trinomial \( (x-1)^2 \). The entire equation now reads: \( P(x) = (x-1)^2 - 16 \), which is in vertex form. Completing the square not only makes it easier to graph but also directly reveals the vertex of the parabola, simplifying analysis of the function.

Parabola Graphing

Graphing a parabola from its vertex form can be done smoothly once you understand the significance of each component. Let's focus on a quick how-to based on the function \(P(x) = (x-1)^2 - 16\):

• Identify the vertex: For this function, it's \( (1, -16) \).
• Determine the direction: The value of \(a\) is 1, signifying the parabola opens upwards.
• Sketch the axis of symmetry: This is a vertical line through the x-coordinate of the vertex, \(x = 1\).
• Choose nearby x-values: Pick points like \(x = 0\) and \(x = 2\) and calculate their corresponding \(P(x)\) values to plot extra points. For example, \(P(0) = 1^2 - 16 = -15\) and \(P(2) = 1^2 - 16 = -15\).

This technique will help in sketching a neat and precise graph. By plotting the vertex and using the symmetry alongside additional calculated points, you can illustrate the beautiful U-shape of the parabola efficiently. The graph should clearly show how it rises from its lowest point, the vertex, symmetrically on both sides.