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Chapter 11: Problem 74

Describe how to find a parabola's vertex if its equation is in the form \(f(x)=a x^{2}+b x+c .\) Use \(f(x)=x^{2}-6 x+8\) as an example.

Short Answer

Expert verified
The vertex of the parabola given by the equation \(f(x)=x^{2}-6x+8\) is at (3, 1).

Step by step solution

01

Identify the coefficients a, b and c

The general form of the equation of a parabola is \(f(x) = a x^{2} + b x + c\). Compare this with the given equation \(f(x)=x^{2}-6x+8\) to identify the coefficients. Here \(a = 1\), \(b = -6\), and \(c = 8\).
02

Calculate the x-coordinate of the vertex

The formula for the x-coordinate is \(h = -\frac{b}{2a}\). Substituting the values of a and b, we get \(h = -\frac{-6}{2*1} = 3\). So, the x-coordinate of the vertex is 3.
03

Calculate the y-coordinate of the vertex

The y-coordinate is found by substituting the x-coordinate into the equation. That is, \(k = f(h) = f(3) = (3)^2 - 6*(3) + 8 = 1 \). So, the y-coordinate of the vertex is 1.

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