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

Jun 6, 2018

$\left(- 1 , - 0.612\right)$

#### Explanation:

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

i.e. $\left(\frac{- b}{2 a} , \frac{- D}{4 a}\right)$ ... For $a {x}^{2} + b x + c = 0$

Here, $D$ is Discriminant which is $= \sqrt{{b}^{2} - 4 a c}$. It also determines the nature of the roots of the equation.

Now, in the given equation ;

$a = 2$

$b = 4$

$c = - 1$

$D = \sqrt{{b}^{2} - 4 a c} = \sqrt{{4}^{2} - 4 \left(2\right) \left(- 1\right)} = \sqrt{16 + 8} = \sqrt{24} = 2 \sqrt{6}$

$\therefore$ By applying the vertex formula here, we get

$\left(\frac{- b}{2 a} , \frac{- D}{4 a}\right) = \left(\frac{- 4}{2 \times 2} , \frac{- 2 \sqrt{6}}{4 \times 2}\right)$

$= \left(\frac{- 4}{4} , \frac{- 2 \sqrt{6}}{8}\right)$

$= \left(- 1 , \frac{- \sqrt{6}}{4}\right)$

$= \left(- 1 , - 0.612\right)$ Hence, the vertex of the equation $f \left(x\right) = 2 {x}^{2} + 4 x - 1 = 0$ is $\left(- 1 , - 0.612\right)$