Problem 30 - Algebra 2 and Trigonometry

Question

In \(28-33,\) without graphing the parabola, describe the translation, reflection, and \(/\) or scaling that must be applied to \(y=x^{2}\) to obtain the graph of each given function. $$ f(x)=x^{2}-6 x-7 $$

Step-by-Step Solution

Answer

Translate right 3 units and down 16 units; no reflection or scaling.

Step 1: Understand the Standard Equation

The standard form of a quadratic function is given by \(y = ax^2 + bx + c\). In our case, we have \(f(x) = x^2 - 6x - 7\), which corresponds to \(a=1\), \(b=-6\), and \(c=-7\).

Step 2: Transformation Identification

The given function \(f(x) = x^2 - 6x - 7\) can be rewritten by completing the square, allowing us to identify translations. Start with \(x^2 - 6x\). Complete the square to transform this part into \((x - 3)^2 - 9\).

Step 3: Completing the Square

Add and subtract \(9\) (which is \((\frac{-6}{2})^2\)) inside the function to form a perfect square trinomial: \(x^2 - 6x = (x-3)^2 - 9\). Therefore, rewrite the function as \(f(x) = (x-3)^2 - 9 - 7\).

Step 4: Simplify the Function

Simplify the expression to \(f(x) = (x-3)^2 - 16\). Now the function is written in vertex form \(f(x) = (x-h)^2 + k\), where the vertex \((h, k)\) indicates the function's horizontal and vertical shifts from the standard parabola \(y = x^2\).

Step 5: Identify Horizontal and Vertical Translations

Identify from the vertex form that the function translates the graph of \(y = x^2\) right by \(3\) units (because of \(x-3\)) and down by \(16\) units (because of \(-16\)).

Step 6: Reflection and Scaling

Analyze the coefficient of \(x^2\). Since \(a=1\), there are no reflections or vertical scaling (compression/stretching) involved, as these are indicated by changes in \(a\).

Key Concepts

Completing the Square

To better understand quadratic transformations, let's start by discussing 'completing the square'. This method allows us to transform a given quadratic equation into a more useful form, particularly the vertex form. The vertex form of a quadratic equation makes it easier to identify graph translations and reflections.

Completing the square involves rewriting a quadratic expression like \(x^2 - 6x\) into a perfect square trinomial. Look at how we do it: we first take the linear coefficient (\(-6\) in this case), halve it, and square it to get \(9\).

Let's apply this:

• The expression becomes \(x^2 - 6x = (x-3)^2 - 9\).
• We subtract \(9\) because when we re-write \(x^2 - 6x\) as \((x-3)^2\), we are inadvertently adding an extra \(9\) that we now need to remove.

This creates a neat square that is easier to manipulate for various transformations. By completing the square, you can convert standard form equations into a more insightful vertex form where you can easily see the vertex of the parabola.

Graph Translations

Once the quadratic is in vertex form, identifying translations becomes straightforward. Translations involve shifting the graph horizontally and/or vertically.

The transformed equation from our example is \(f(x) = (x-3)^2 - 16\). This showcases how the graph of \(y = x^2\) transforms:

• The term \((x-3)\) suggests a horizontal shift to the right by 3 units. This happens because the equation takes the form \((x-h)^2\), moving the vertex horizontally.
• The constant \(-16\) indicates a vertical shift down by 16 units. It affects the y-coordinate of the vertex, shifting the entire graph vertically.

Graph translations help in moving the basic parabola without changing its shape. By simply adding or subtracting numbers in the vertex form, you can achieve this effortlessly.

Vertex Form of Quadratic

The vertex form of a quadratic equation is essential in understanding and visualizing the graph's translations and vertex. The form is given by \(f(x) = a(x-h)^2 + k\), where \((h, k)\) represents the vertex of the parabola.

In our function, \(f(x) = (x-3)^2 - 16\), this is already in vertex form. Here, we can easily deduce:

• The vertex \((h, k)\) of this parabola is \((3, -16)\).
• This tells us that the parabola is centered 3 units right from the origin along the x-axis and 16 units down along the y-axis.
• The coefficient \(a = 1\) indicates that the parabola opens upwards with the same width as \(y = x^2\).

Understanding the vertex form makes it simpler to visualize the graph and any transformations it undergoes. By examining the vertex and any shifts indicated in the equation, we can map the graph's changes and anticipate its position on the coordinate plane.