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

Problem

ƒ(x)=x^2−2*x−8

Solution

1. Identify the coefficients of the quadratic function in the form ƒ(x)=a*x^2+b*x+c Here, a=1 b=−2 and c=−8

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

x=−(−2)/(2*(1))

x=1

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

ƒ(1)=(1)^2−2*(1)−8

ƒ(1)=1−2−8

ƒ(1)=−9

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

Final Answer

Vertex=(1,−9)

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