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Chapter 0: Problem 30

Show that the vertex of the parabola \(f(x)=a x^{2}+b x+c\) where \(a \neq 0\), is \((-b /(2 a), f(-b /(2 a)))\).

Short Answer

Expert verified
The vertex of the parabola \(f(x) = ax^2 + bx + c\) is \(\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)\) by completing the square and identifying the vertex coordinates in the new quadratic function.

Step by step solution

01

Write the quadratic function in standard form

Before we complete the square, let's first write our given quadratic function \(f(x) = ax^2 + bx + c\) explicitly, with a, b, and c being constants: \(f(x) = ax^2 + bx + c\)
02

Complete the square

Now, we will complete the square to find the new expression for the quadratic function: Notice that, in order to complete the square, we need to rewrite \(ax^2 + bx\) such that a perfect square is formed. The first step to do this is to factor out the leading coefficient \(a\) from \(ax^2 + bx\): \(f(x) = a(x^2 + \frac{b}{a}x)+ c\) Next, we will add and subtract the square of half of the x coefficient inside the parenthesis: \(f(x) = a\left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right) + c\) \(f(x) = a\left(\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right) + c\) Now, we have completed the square.
03

Simplify the expression

Let's distribute and simplify the expression to get our new quadratic function: \(f(x) = a\left(x + \frac{b}{2a}\right)^2 - a\left(\frac{b}{2a}\right)^2 + c\)
04

Identify the vertex

Since our given quadratic function is now in the form \(f(x) = a\left(x + \frac{b}{2a}\right)^2 - a\left(\frac{b}{2a}\right)^2 + c\), we can identify the vertex of the parabola. The vertex is given by the point \((h, k)\), where h is the opposite of the term inside the parentheses and k is the value of the function at that point: \(h = -\frac{b}{2a}\) To find the vertex's y-coordinate, we substitute h back into the function: \(k = f(h) = a\left(-\frac{b}{2a} + \frac{b}{2a}\right)^2 - a\left(\frac{b}{2a}\right)^2 + c\) Since the first term becomes zero, we have: \(k = f(h) = - a\left(\frac{b}{2a}\right)^2 + c\)
05

Write the vertex coordinates

The vertex of the parabola is given by the point \((h, k)\), which is: \(\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)\) Thus, we have demonstrated that the vertex of the parabola \(f(x) = ax^2 + bx + c\) is indeed \((-\frac{b}{2a}, f(-\frac{b}{2a}))\).

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