Problem 63 - College Algebra and Calculus: An Applied Approach

Question

Write the quadratic function \(f(x)=a x^{2}+b x+c\) in standard form to verify that the vertex occurs at \(\left(-\frac{b}{2 a}, f\left(-\frac{b}{2 a}\right)\right)\)

Step-by-Step Solution

Answer

The vertex in the standard form of the quadratic function corresponds to the vertex \(\left(-\frac{b}{2 a}, f\left(-\frac{b}{2 a}\right)\right)\) in the general form.

Step 1: Write the formula for standard form of quadratic equation

Quadratic function can be rewritten in standard form as \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the function.

Step 2: Equate coefficient from general form to standard form

In the general form equation \(f(x) = ax^{2} + bx + c\), equate it to the standard form equation: \[ a(x-h)^2 + k = ax^{2} + bx + c \] Expand the left hand side (LHS): \[ a(x^2 - 2hx + h^2) + k = ax^{2} + bx + c \] By comparing the coefficients, the following equations are obtained: \[ -2ah = b \] \[ ah^2 + k = c \]

Step 3: Calculate the vertex using h and k

To find the vertex, first, solve for h: \( h = -b/2a \) Second, substitute h into 'ah^2 + k = c' to solve for k: \( k = f(h) = c - ah^2 = c - a(-b/2a)^2 = c - b^2/4a = f(-b/2a)\) The vertex is (h, k) = \(\left(-\frac{b}{2 a}, f\left(-\frac{b}{2 a}\right)\right)\)

Key Concepts

Vertex Form

The vertex form of a quadratic function is a way to express the equation so that we can easily identify the vertex of the parabola it represents. The formula for the vertex form is \(f(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex. By having the vertex form, it's simple to see the turning point of the graph.
The "\(a\)" in front is the same coefficient as in the regular standard form, controlling the parabola's direction and width.
When the value of \"a\" is positive, the parabola opens upwards; when negative, it opens downwards.
This form is invaluable when analyzing the graph, as slight changes in \((h, k)\) directly shift the parabola's position. Here’s how you can visualize each part:

• "\(h\)" moves the vertex along the x-axis.
• "\(k\)" moves it along the y-axis.
• The shape remains consistent unless "\(a\)" changes.

Understanding the vertex form makes it simpler to graph and comprehend a quadratic equation.

Standard Form

The standard form of a quadratic equation is the one most students first learn, expressed as \(f(x) = ax^2 + bx + c\). Each term plays a distinct role in shaping the parabola:
The "\(a\)" value affects the direction (up or down) and the width of the parabola; if \(a\) is zero, it’s not a quadratic.
The "\(b\)" and "\(c\)" terms further influence the graph’s position. Let's break down the components:

• "\(a\)": Determines direction (up for positive, down for negative).
• "\(b\)": Influences the steepness and horizontal position.
• "\(c\)": Represents where the parabola crosses the y-axis.

This form is straightforward for identifying the y-intercept directly.
However, identifying the vertex is less obvious compared to the vertex form.
To understand the vertex from this form, conversion or use of the vertex formula \((-b/2a, f(-b/2a))\) is necessary.

Completing the Square

Completing the square is a method used to convert the standard form of a quadratic equation into the vertex form. This process helps in revealing the vertex easily, making the function more manageable. Here's a simple breakdown of how the procedure works:
Let's consider the quadratic function \( f(x) = ax^2 + bx + c \).

• Step 1: Factor out the "\(a\)" from the first two terms if \(aeq1\), so that the equation looks like \( a(x^2 + \frac{b}{a}x) + c \).
• Step 2: Take \(\frac{b}{2a}\), square it, and add/subtract inside the bracket to complete the square: \( a(x^2 + \frac{b}{a}x + (\frac{b}{2a})^2 - (\frac{b}{2a})^2) + c \).
• Step 3: Rearrange so you have a perfect square trinomial inside the bracket: \( a((x + \frac{b}{2a})^2 - (\frac{b}{2a})^2) + c \).
• Step 4: Simplify to achieve the vertex form: \( a(x + \frac{b}{2a})^2 + (c - a(\frac{b}{2a})^2) \).

This method shows how the vertex \((h, k)\) is derived clearly, as the equation now reads \(a(x-h)^2 + k\).
It's a powerful way to visualize where the function's vertex is and graph the parabola more effectively.