What is the vertices f(x)=-2x^2-x+1 ?

The vertices f(x)=-2x^2-x+1 is Maximum (-1/4 , 9/8)

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vertices f(x)=-2x^2-x+1

vertices

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

Maximum (- \frac{1}{4}, \frac{9}{8})

Solution steps

y = - 2 x^{2} - x + 1

The parabola parameters are:

a = - 2, b = - 1, c = 1

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

x_{v} = - \frac{( - 1 )}{2 ( - 2 )}

Simplify

x_{v} = - \frac{1}{4}

Plug in

x_{v} = - \frac{1}{4}

to find the

y_{v}

value

y_{v} = \frac{9}{8}

Therefore the parabola vertex is

(- \frac{1}{4}, \frac{9}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (- \frac{1}{4}, \frac{9}{8})

- x^{2} + 3 x + 1

f (x) = x^{2} + 12 x + 35

y^{2} + 14 y + 4 x + 45 = 0

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

x^{2} + 8 x + 12

What is the vertices f(x)=-2x^2-x+1 ?
The vertices f(x)=-2x^2-x+1 is Maximum (-1/4 , 9/8)