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 Here, 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=(−6)/(−2)

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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