Find the Vertex f(x)=-x^2-6x+3

Problem

ƒ(x)=−x2−6*x+3

Solution

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

a=−1

b=−6

c=3

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.

ƒ*(−3)=−(−3)2−6*(−3)+3

ƒ*(−3)=−(9)+18+3

ƒ*(−3)=12

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

Vertex=(−3,12)

Final Answer

Vertex of *ƒ(x)=−x2−6*x+3* is *(−3,12)

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