Problem 29 - Beginning and Intermediate Algebra

Question

Rewrite function in the form \(f(x)=a(x-h)^{2}+k\) by completing the square. Then, graph the function. Include the intercepts. \(y=x^{2}+6 x+7\)

Step-by-Step Solution

Answer

The given function \(y = x^2 + 6x + 7\) can be rewritten in the form \(f(x) = a(x-h)^2 + k\) as \(f(x) = (x+3)^2 - 2\). The vertex is \((-3, -2)\), the axis of symmetry is \(x = -3\), the x-intercepts are \((-3 + \sqrt{2}, 0)\) and \((-3 - \sqrt{2}, 0)\), and the y-intercept is \((0, 7)\). To graph the function, plot these points and the axis of symmetry, then sketch the parabola symmetrically around the axis of symmetry.

Step 1: Identify the quadratic and linear terms of the function

The given function is \(y = x^2 + 6x + 7\). The quadratic term is \(x^2\), and the linear term is \(6x\).

Step 2: Find the square completion term

To complete the square, we need to find the value that makes the quadratic and linear terms form a perfect square trinomial. We can do this with the following formula: \(\frac{b}{2}\)^2, where \(b\) is the coefficient of the linear term. In this case, \(b = 6\). Calculating the square completion term: \(\left(\frac{6}{2}\right)^2 = 3^2 = 9\)

Step 3: Rewrite the function by completing the square

We add and subtract the square completion term, 9, to the original function and then factor it. \(y = x^2 + 6x + 7\) \(y = (x^2 + 6x + 9) + 7 - 9\) \(y = (x + 3)^2 - 2\) So the function rewritten in the form \(f(x)=a(x-h)^2+k\) is \(f(x) = (x + 3)^2 - 2\).

Step 4: Identify the vertex and axis of symmetry

From the rewritten function, we can identify the vertex and axis of symmetry. The vertex is given by the coordinates \((-h, k)\), so in this case, the vertex is \((-3, -2)\). The axis of symmetry is the vertical line given by the equation \(x = -3\).

Step 5: Find the intercepts

1. To find the x-intercept(s), we set \(y = 0\), and solve for x: \(0 = (x + 3)^2 - 2\) \(2 = (x + 3)^2\) \(x + 3 = \pm\sqrt{2}\) \(x = -3 \pm\sqrt{2}\) So the x-intercepts are \((-3 + \sqrt{2}, 0)\) and \((-3 - \sqrt{2}, 0)\). 2. To find the y-intercept, we set \(x = 0\), and solve for y: \(y = (0 + 3)^2 - 2\) \(y = 7\) So the y-intercept is \((0, 7)\).

Step 6: Graph the function

1. Plot the vertex \((-3, -2)\), the x-intercepts \((-3 + \sqrt{2}, 0)\) and \((-3 - \sqrt{2}, 0)\), and the y-intercept \((0, 7)\). 2. Draw the axis of symmetry, the vertical line at \(x = -3\). 3. Sketch the parabola using these points, making sure it is symmetric about the axis of symmetry.

Key Concepts

Vertex Form

The purpose of rewriting a quadratic function in vertex form is to make it easy to understand its characteristics, especially the vertex of the parabola. The vertex form of a quadratic function is given by \[ f(x) = a(x-h)^2 + k \] where:

• a is the coefficient that affects the width and direction of the parabola.
• (h, k) represents the coordinates of the vertex.

The vertex form is particularly helpful for graphing, as it immediately tells us where the parabola is centered on the coordinate plane. In the example from the exercise, the quadratic function \( y = (x + 3)^2 - 2 \) is in vertex form. This directly tells us that:- The vertex is \( (-3, -2) \).- The axis of symmetry is the vertical line \( x = -3 \).Understanding these elements allows us to easily construct a graph without having to calculate additional points first.

Quadratic Functions

Quadratic functions are polynomial functions of the second degree. They take the general form:\[ y = ax^2 + bx + c \] where:

• a, b, and c are constants.
• The graph of a quadratic function is always a parabola.
• The parabola opens upwards if a is positive and downwards if it's negative.

In the quadratic function \( y = x^2 + 6x + 7 \), the term \( x^2 \) makes the parabola open upwards because the coefficient of \( x^2 \) (which is 1) is positive. Completing the square is a technique frequently used to convert any quadratic function into vertex form, simplifying the process of analyzing the function's properties.

Graphing Parabolas

Graphing a parabola correctly involves identifying key points and characteristics. Here’s how to graph using vertex form:

• Identify the Vertex: The vertex is the central turning point of the parabola. In our function \( y = (x + 3)^2 - 2 \), the vertex is given by \( (-3, -2) \).


• Axis of Symmetry: A vertical line running through the vertex, it equally divides the parabola. For our function, it is \( x = -3 \).


• Plot Intercepts: Calculating and plotting x and y-intercepts adds detailed reference points to the graph.


• Draw the Parabola: Starting from the vertex, sketch the curve equally distant from the axis of symmetry, ensuring it captures the parabolic nature.

Employ these steps to visualize how the function behaves on the graph, revealing its symmetry and shape.

Intercepts Calculation

Intercepts are crucial points where the graph intersects the axes, providing insight into the function's behavior. There are two types of intercepts for quadratic functions:

• X-intercepts (Roots): Found by setting \( y = 0 \) and solving for \( x \). For our function \( (x + 3)^2 - 2 = 0 \), solving gives: \[ x = -3 \pm \sqrt{2} \] resulting in intercepts at \( (-3 + \sqrt{2}, 0) \) and \( (-3 - \sqrt{2}, 0) \).


• Y-intercept: Found by setting \( x = 0 \) and solving for \( y \). Substituting gives \( y = 7 \), resulting in the point \( (0, 7) \).

These intercepts are integral for plotting the parabola on the coordinate grid and give a clear picture of where the graph crosses the axes.