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

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Answer 1 · 2018-06-06T12:00:43

#(-1 , -0.612)#

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)#