Question

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

Answer 1

`(-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)`


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