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

Problem

ƒ(x)=2*x2+x−6

Solution

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

a=2

b=1

c=−6

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

x=−1/(2*(2))

x=−1/4

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

y=ƒ*(−1/4)

y=2*(−1/4)2+(−1/4)−6

4. Simplify the expression to determine the value of y

y=2*(1/16)−1/4−6

y=1/8−2/8−48/8

y=−49/8

Final Answer

Vertex=(−1/4,−49/8)

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