Find the Vertex f(x)=3x^2+5x-4

Problem

ƒ(x)=3*x2+5*x−4

Solution

1. Identify the coefficients of the quadratic function in the form ƒ(x)=a*x2+b*x+c where a=3 b=5 and c=−4

2. Apply the formula for the xcoordinate of the vertex, which is given by x=−b/(2*a)

3. Substitute the values of a and b into the formula to find the xcoordinate.

x=−5/(2*(3))

x=−5/6

4. Evaluate the function at x=−5/6 to find the ycoordinate of the vertex.

ƒ*(−5/6)=3*(−5/6)2+5*(−5/6)−4

ƒ*(−5/6)=3*(25/36)−25/6−4

ƒ*(−5/6)=25/12−50/12−48/12

ƒ*(−5/6)=−73/12

Final Answer

Vertex=(−5/6,−73/12)

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