Problem 9 - College Algebra with Corequisite Support

Question

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

Step-by-Step Solution

Answer

The standard form is \( f(x) = \left(x + \frac{5}{2}\right)^2 - \frac{33}{4} \) with the vertex at \( \left(-\frac{5}{2}, -\frac{33}{4}\right) \).

Step 1: Identify the Quadratic Equation

The given quadratic function is \( f(x) = x^2 + 5x - 2 \). We need to rewrite it in the vertex form, which is \( f(x) = a(x-h)^2 + k \), and identify the vertex \((h, k)\).

Step 2: Complete the Square

To complete the square, rewrite the equation with the terms needed for completing the square. Start with the structure: \( x^2 + 5x \). Add and subtract \( \left(\frac{5}{2}\right)^2 \) inside the function: \[ x^2 + 5x = \left(x + \frac{5}{2}\right)^2 - \left(\frac{5}{2}\right)^2 \]},

Step 3: Simplify the Completed Square

Now express the entire quadratic function using the completed square:\[ f(x) = \left(x + \frac{5}{2}\right)^2 - \frac{25}{4} - 2 \]Simplify the expression:\[ f(x) = \left(x + \frac{5}{2}\right)^2 - \frac{25}{4} - \frac{8}{4} \ = \left(x + \frac{5}{2}\right)^2 - \frac{33}{4} \]

Step 4: Extract the Vertex

The quadratic function is now in the form \((x + \frac{5}{2})^2 - \frac{33}{4}\). From this, identify the vertex \((h, k)\), where\[ h = -\frac{5}{2}, \ k = -\frac{33}{4} \]. Thus, the vertex is \( \left(-\frac{5}{2}, -\frac{33}{4}\right) \).

Key Concepts

Completing the Square

Completing the square is a method used in algebra to rewrite a quadratic equation in a form that makes it easier to solve or analyze. This process turns a standard quadratic equation like \( f(x) = x^2 + 5x - 2 \) into a 'perfect square trinomial.' This trinomial can then be expressed as the square of a binomial, which simplifies finding the vertex.
To complete the square for the given equation, first focus on the quadratic and linear terms, \( x^2 + 5x \). We need to find a constant that allows us to express these as a squared binomial expression. Here, we take half of the coefficient of \( x \) (which is 5), and square it: \( \left( \frac{5}{2} \right)^2 = \frac{25}{4} \).
Then, add and subtract this constant within the equation to maintain balance:

• Rewrite: \( x^2 + 5x = \left(x + \frac{5}{2}\right)^2 - \frac{25}{4} \).
• Include the rest of the original function to get: \( f(x) = \left(x + \frac{5}{2}\right)^2 - \frac{25}{4} - 2 \).

Vertex Form

The vertex form of a quadratic function is a way of expressing the equation that highlights the vertex, or the highest or lowest point of the parabola. This form is written as \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola.
By converting a quadratic equation into vertex form, you make it straightforward to identify the vertex, since the values \( h \) and \( k \) are present directly in the equation.
For the example \( f(x) = \left(x + \frac{5}{2}\right)^2 - \frac{33}{4} \), by completing the square, we have expressed the quadratic in vertex form, where:

• \( h = -\frac{5}{2} \)
• \( k = -\frac{33}{4} \)

The vertex form not only allows us to see the vertex readily but provides a basis for graphing the parabola easily, as the vertex \((h, k)\) indicates the turning point and the parameter \( a \) shows the direction of the parabola.

Vertex Calculation

Calculating the vertex of a quadratic function is crucial for understanding its graphical representation. Once a quadratic is in vertex form \( f(x) = a(x-h)^2 + k \), finding the vertex becomes a matter of identifying \( h \) and \( k \) directly from the equation.
In our provided solution, after completing the square, the function is restructured into the vertex form \( f(x) = \left(x + \frac{5}{2}\right)^2 - \frac{33}{4} \).

• This directly informs us that the vertex \( (h, k) \) is \( \left(-\frac{5}{2}, -\frac{33}{4}\right) \).

Understanding vertex calculation helps in sketching the curve of the quadratic accurately, as the vertex signifies where the function achieves its minimum or maximum value depending on the sign of the leading coefficient \( a \). If \( a \) is positive, the parabola opens upwards, and if negative, it opens downwards.