What is the vertex of f(x)=2x^2+4x-1 ?

1 Answer
Jun 6, 2018

(-1 , -0.612)

Explanation:

To solve this question, we need to know the formula for finding vertex of a general equation.

i.e. ((-b)/(2a) , (-D)/(4a)) ... For ax^2+bx+c=0

Here, D is Discriminant which is =sqrt(b^2-4ac). It also determines the nature of the roots of the equation.

Now, in the given equation ;

a = 2

b = 4

c =-1

D=sqrt(b^2-4ac)=sqrt(4^2-4(2)(-1))=sqrt(16+8)=sqrt24=2sqrt6

:. By applying the vertex formula here, we get

((-b)/(2a) , (-D)/(4a))=((-4)/(2xx2) , (-2sqrt6)/(4xx2))

=((-4)/(4) , (-2sqrt6)/(8))

=(-1 , (-sqrt6)/4)

=(-1 , -0.612)

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Hence, the vertex of the equation f(x)=2x^2+4x-1=0 is (-1 , -0.612)