Find the Vertex f(x)=x^2+2x-3

Problem

ƒ(x)=x^2+2*x−3

Solution

1. Identify the coefficients of the quadratic function in the form ƒ(x)=a*x^2+b*x+c where a=1 b=2 and c=−3

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)−3

ƒ*(−1)=1−2−3

ƒ*(−1)=−4

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

Final Answer

Vertex=(−1,−4)

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