Find the Vertex f(x)=3x^2+18x+29

Problem

ƒ(x)=3*x2+18*x+29

Solution

1. Identify the coefficients of the quadratic function in the form ƒ(x)=a*x2+b*x+c where a=3 b=18 and c=29

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

x=−18/(2*(3))

x=−18/6

x=−3

3. Substitute the xcoordinate back into the original function to find the ycoordinate, ƒ*(−3)

ƒ*(−3)=3*(−3)2+18*(−3)+29

ƒ*(−3)=3*(9)−54+29

ƒ*(−3)=27−54+29

ƒ*(−3)=2

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

Vertex=(−3,2)

Final Answer

Vertex of *ƒ(x)=3*x2+18*x+29* is *(−3,2)

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