Problem 66 - Precalculus

Question

Rewrite the quadratic functions in standard form and give the vertex. \(f(x)=x^{2}-x\)

Step-by-Step Solution

Answer

The standard form is \( f(x) = (x - \frac{1}{2})^2 - \frac{1}{4} \). The vertex is \( \left( \frac{1}{2}, -\frac{1}{4} \right) \).

Step 1: Identify the standard form of a quadratic function

The standard form of a quadratic function is \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola. Our goal is to rewrite the given function in this form.

Step 2: Complete the square

To convert \(f(x) = x^2 - x\) to standard form, we need to complete the square. Start by rearranging the expression: \(x^2 - x = (x^2 - x + \frac{1}{4}) - \frac{1}{4}\). We add and subtract \(\frac{1}{4}\) to complete the square.

Step 3: Rewrite the expression as a perfect square

Now, notice that \(x^2 - x + \frac{1}{4}\) can be rewritten as \((x - \frac{1}{2})^2\). This is because \((x - \frac{1}{2})^2\) expands to \(x^2 - x + \frac{1}{4}\). Thus, the function becomes \(f(x) = (x - \frac{1}{2})^2 - \frac{1}{4}\).

Step 4: Identify the vertex from the standard form

From the expression \((x - \frac{1}{2})^2 - \frac{1}{4}\), we can identify the vertex \((h, k)\) as \(\left(\frac{1}{2}, -\frac{1}{4}\right)\).

Key Concepts

Standard Form

Quadratic functions are often expressed in a format called the standard form. This form is particularly useful because it makes it easier to understand and visualize the properties of the function, such as its vertex, axis of symmetry, and the direction in which its parabola opens. Standard form is written as \( f(x) = a(x-h)^2 + k \), where:

• \(a\) is a coefficient that affects the width and direction of the parabola.
• \(h\) and \(k\) are constants that determine the location of the vertex on the graph.

An important thing to remember is that \((h,k)\) becomes the vertex of the parabola. The standard form is particularly advantageous when you want to easily identify the vertex and sketch the graph of the quadratic function.

Vertex

The vertex of a quadratic function is a key feature that gives valuable insights into the function's graph. It's the highest or lowest point on the parabola, depending on whether it opens upwards or downwards. Here's how you can find the vertex from both standard form and general form:

• If the quadratic is expressed in standard form \(f(x) = a(x-h)^2 + k\), the vertex is simply \((h, k)\).
• For a quadratic in general form \(ax^2 + bx + c\), you can calculate the vertex using the formula \(h = -\frac{b}{2a}\), and then find \(k\) by substituting \(h\) back into the equation.

In context, the vertex \(\left(\frac{1}{2}, -\frac{1}{4}\right)\) was determined after rewriting the quadratic \(x^2 - x\) into its standard form.

Completing the Square

Completing the square is a crucial algebraic process used to transform a quadratic function into its standard form. This technique is not only useful for converting forms but also for solving quadratic equations and graphing parabolas. Here’s a simplified breakdown:

• You begin by rearranging the quadratic terms to focus on making a perfect square trinomial.
• Take the coefficient of \(x\), divide by two, and square it. This value is added and subtracted within the expression to maintain balance.

For example, in \(x^2 - x\), we add and subtract \(\frac{1}{4}\) to achieve \((x - \frac{1}{2})^2 - \frac{1}{4}\), which represents the standard form. Completing the square hence reveals the vertex form, enabling clearer insights into the function's properties.