Find the Vertex f(x)=x^2-6x+8

Problem

ƒ(x)=x2−6*x+8

Solution

1. Identify the coefficients of the quadratic function in the form ƒ(x)=a*x2+b*x+c where a=1 b=−6 and c=8

2. Calculate the xcoordinate of the vertex using the formula x=(−b)/(2*a)

x=(−(−6))/(2*(1))

x=3

3. Substitute the xcoordinate back into the original function to find the ycoordinate of the vertex.

ƒ(3)=(3)2−6*(3)+8

ƒ(3)=9−18+8

ƒ(3)=−1

4. State the vertex as the ordered pair (x,y)

Final Answer

Vertex=(3,−1)

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