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Chapter 3: Problem 78

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
Vertex of the parabola \(f(x) = x^{2}-6x+8\) is at (3, -1).

Step by step solution

01

Identify the Coefficients

In the given standard form of the parabola, identify the coefficients a, b, and c for the equation \(f(x) = x^{2}-6x+8\), where a is the coefficient of \(x^{2}\), b is the coefficient of x, and c is the constant. Thus, here, a = 1, b = -6 and c = 8.
02

Calculate the x-coordinate of the Vertex

Compute the value of h, which is the x-coordinate of the vertex. The formula to find h is \(-b/2a\). Substituting the values we get, h = \(-(-6)/2*(1) = 3\).
03

Calculate the y-coordinate of the Vertex

Compute the value of k, which is the y-coordinate of the vertex. The formula to find k is \(f(h)\), substitute the value of h in the function gets us k = \(f(3) = (3)^{2} - 6*(3) + 8 = -1\).
04

Identify the Vertex

The vertex of a parabola is represented as (h, k). Here h = 3 and k = -1. So, the vertex of the given parabola is (3, -1).

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