Problem 13 - College Algebra

Question

For the following exercises, rewrite the quadratic functions in standard form and give the vertex. $$ f(x)=3 x^{2}-5 x-1 $$

Step-by-Step Solution

Answer

Vertex: \((\frac{5}{6}, -\frac{37}{12})\)

Step 1: Identify the Objective

We need to rewrite the quadratic function in its standard form and find the vertex of the function. The given function is \( f(x)=3x^2-5x-1 \).

Step 2: Recall Standard Form

The standard form of a quadratic function is \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola.

Step 3: Complete the Square

We start by completing the square for the quadratic expression.\[ f(x) = 3(x^2 - \frac{5}{3}x) - 1 \]To complete the square, find the number that makes the trinomial a perfect square. Take \(-\frac{5}{3}\), halve it to get \(-\frac{5}{6}\), and square it to get \(\frac{25}{36}\).Add and subtract \(3\times\frac{25}{36}\) inside the parentheses.\[ f(x) = 3 \left( x^2 - \frac{5}{3}x + \frac{25}{36} - \frac{25}{36}\right) - 1 \]\[ f(x) = 3 \left( (x-\frac{5}{6})^2 - \frac{25}{36} \right) - 1 \]Distribute 3:\[ f(x) = 3(x-\frac{5}{6})^2 - 3 \times \frac{25}{36} - 1 \]\[ f(x) = 3(x-\frac{5}{6})^2 - \frac{25}{12} - 1 \] \[ f(x) = 3(x-\frac{5}{6})^2 - \frac{37}{12} \]

Step 4: Identify the Vertex

The expression now is in standard form \[ f(x) = 3(x-\frac{5}{6})^2 - \frac{37}{12} \] Thus, the vertex \((h, k)\) is \((\frac{5}{6}, -\frac{37}{12})\).

Key Concepts

Standard Form

When working with quadratic functions, the standard form is an essential representation. The standard form of a quadratic function is given by:\[ f(x) = a(x-h)^2 + k \]where \(a\) is a constant and \( (h, k) \) is the vertex of the parabola. This form makes it easy to identify the vertex, which is a key feature of any quadratic function.

The standard form allows us to see the shift of the parabola from its basic position at the origin. This makes it very useful when we need to graph the function or analyze its behavior, such as its maximum or minimum point. Converting a quadratic equation into standard form often involves completing the square, which we'll explore next.

Understanding standard form can also help us determine some important characteristics of a parabola:

• Direction of opening (upward or downward) depending on the sign of \(a\).
• The vertex, which gives us the minimum or maximum value.
• Axis of symmetry, a vertical line x = \(h\).

Completing the Square

Completing the square is a technique used to transform a quadratic equation into its standard form. This is done by rewriting the quadratic part of the equation as a perfect square trinomial.

Let's take the example function, \[ f(x)=3x^2-5x-1 \]The first step in completing the square is to factor out the coefficient of the \(x^2\) term, so we can focus on the quadratic and linear terms:
\[ f(x) = 3(x^2 - \frac{5}{3}x) - 1 \]Next, we make the quadratic expression inside the parentheses a perfect square trinomial. To do this, take half of the \(x\) term's coefficient and square it.

• Halve \( -\frac{5}{3} \) to get \( -\frac{5}{6} \).
• Square it to obtain \( \frac{25}{36} \).

Add and subtract this squared value inside the parentheses, then factor the trinomial:\[ f(x) = 3((x - \frac{5}{6})^2 - \frac{25}{36}) - 1 \]This new expression highlights the perfect square. Finally, distribute any factors and simplify the constant terms to finish converting into standard form.

Vertex of a Parabola

After converting a quadratic to its standard form, the vertex of the parabola becomes easily identifiable. The vertex is a significant feature of the parabola, indicating the peak or the trough of the curve.

In the standard form \( f(x) = a(x-h)^2 + k \), the vertex is located at \((h, k)\). For the example equation that we worked with, \[ f(x) = 3(x-\frac{5}{6})^2 - \frac{37}{12} \]The vertex is at:

• \( h = \frac{5}{6} \)
• \( k = -\frac{37}{12} \)

Thus, the point \( (\frac{5}{6}, -\frac{37}{12}) \) represents the vertex. This point is crucial as it helps in:

• Understanding the maximum or minimum value of the function (depending on whether the parabola opens upwards or downwards).
• Analyzing symmetry since the vertex also lies on the axis of symmetry of the parabola.
• Determining the function's rate of change in different intervals.

Grasping the concept of the vertex aids in graphing the quadratic function effectively and understanding its real-world applications.