What is the vertices f(x)=-2x^2-4x+3 ?

The vertices f(x)=-2x^2-4x+3 is Maximum (-1,5)

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vertices f(x)=-2x^2-4x+3

vertices

f (x) = - 2 x^{2} - 4 x + 3

Maximum (- 1, 5)

Solution steps

y = - 2 x^{2} - 4 x + 3

The parabola parameters are:

a = - 2, b = - 4, c = 3

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{( - 4 )}{2 ( - 2 )}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(- 1, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (- 1, 5)

f (x) = - (x^{2} + 7 x - 8)

- x^{2} + 2 x - 3

y = x^{2} + 2 x - 2

y = x^{2} - 3 x^{2}

f (x) = 8 x^{2} + 4 x - 1

What is the vertices f(x)=-2x^2-4x+3 ?
The vertices f(x)=-2x^2-4x+3 is Maximum (-1,5)