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

Problem

ƒ(x)=x2+2*x−8

Solution

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

a=1

b=2

c=−8

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

x=(−2)/(2*(1))

x=−1

3. Calculate the y-coordinate by substituting the x-coordinate back into the original function ƒ(x)

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)

Vertex=(−1,−9)

Final Answer

Vertex of *ƒ(x)=x2+2*x−8* is *(−1,−9)

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