Find the Vertex f(x)=x^2+4x-5

Problem

ƒ(x)=x2+4*x−5

Solution

1. Identify the coefficients of the quadratic function in the form ƒ(x)=a*x2+b*x+c where a=1 b=4 and c=−5

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

x=(−4)/(2*(1))

x=−2

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

ƒ*(−2)=(−2)2+4*(−2)−5

ƒ*(−2)=4−8−5

ƒ*(−2)=−9

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

Final Answer

Vertex=(−2,−9)

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