Problem 6 - 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) Vertex form is \( P(x) = (x+1)^2 - 16 \). (b) The vertex is \( (-1, -16) \). (c) Graph the parabola with vertex \((-1, -16)\) opening upwards.

Step 1: Write the function in vertex form

We start with the given quadratic function: \[ P(x) = x^2 + 2x - 15 \]. To write it in the vertex form \( P(x) = a(x-h)^2 + k \), we complete the square.First, focus on the quadratic and linear terms: \( x^2 + 2x \). To complete the square, we take the coefficient of \( x \), which is 2, divide it by 2 to get 1, and then square it to get 1. Now, add and subtract 1 inside the equation: \[P(x) = (x^2 + 2x + 1) - 1 - 15 \]\[= (x + 1)^2 - 16 \]. This is the vertex form: \( a(x-h)^2+k \) where \( a = 1 \), \( h = -1 \), and \( k = -16 \).

Step 2: Identify the Vertex of the Parabola

With the function in vertex form \( P(x) = (x + 1)^2 - 16 \), the vertex \((h, k)\) can be directly identified as \(h = -1\) and \(k = -16\). Hence, the vertex of the parabola is \((-1, -16)\).

Step 3: Graph the Function

To graph \( P(x) = (x + 1)^2 - 16 \), start by plotting the vertex \((-1, -16)\) on the coordinate plane. The parabola opens upwards since the coefficient \( a = 1 \) is positive.Next, identify additional points to sketch: - For \( x = 0 \), compute \( P(0) = (0+1)^2 - 16 = -15 \). - For \( x = -2 \), compute \( P(-2) = ((-2)+1)^2 - 16 = -15 \).These points, along with symmetry about the line \( x = -1 \), help in sketching the parabola.

Key Concepts

Completing the Square

Completing the square is a technique used in algebra to rewrite a quadratic function into a more workable form. This form helps in easily identifying important characteristics about the function, such as the vertex of its graph. To complete the square, focus only on the terms involving the variable, which in this case are the quadratic term and the linear term.
For the quadratic function \(P(x) = x^2 + 2x - 15\), only consider \(x^2 + 2x\). The steps for completing the square are as follows:

• Identify the coefficient of the linear term (here, the '2' from \(2x\)). Divide this coefficient by 2, resulting in 1.
• Square this result: \(1^2 = 1\).

After these steps, add and subtract the square (1) from the expression, essentially changing nothing as you are adding zero: \(x^2 + 2x + 1 - 1 - 15\). The term \(x^2 + 2x + 1\) simplifies to \((x+1)^2\). Hence, the expression becomes \((x+1)^2 - 16\), which is the quadratic expression in its completed square form.

Vertex Form

The vertex form of a quadratic function provides a simple way to determine the vertex of its graph—a critical point that represents the maximum or minimum value, depending on the direction the parabolaopens. The vertex form is expressed as \(P(x) = a(x-h)^2 + k\). In this format:

• \(a\) indicates the parabola's direction (upwards if positive, downwards if negative).
• \(h\) and \(k\) give the vertex's coordinates, represented as \((h, k)\).

From our exercise, converting \(P(x) = x^2 + 2x - 15\) to vertex form leads to \(P(x) = (x+1)^2 - 16\). Thus, the vertex coordinates from this expression are \(h = -1\) and \(k = -16\), or \((-1, -16)\). Recognizing the vertex form helps in understanding the graph's shape and position without extensive calculations.

Graphing Parabolas

Graphing a parabola from its quadratic function can initially seem complex, but with the vertex form, it becomes much easier. Begin by identifying the vertex, which dictates the starting point on the graph. For \(P(x) = (x+1)^2 - 16\), the vertex is \((-1, -16)\). Mark this point on the coordinate plane.
A positive coefficient \(a = 1\) means the parabola opens upwards.
To fully sketch the parabola, find another couple of points its path passes through by choosing values of \(x\) nearby and calculating the corresponding \(P(x)\) values. This helps in revealing the parabola's width and curvature.

• For \(x=0\), calculate: \(P(0) = (0+1)^2 - 16 = -15\).
• For \(x=-2\), calculate: \(P(-2) = ((-2)+1)^2 - 16 = -15\).

These symmetric points about the line \(x = -1\) show the curve's symmetry and help to sketch the parabola accurately, illustrating its nature and direction on the graph.