What is the vertices f(x)=-2x^2-3x+1 ?

The vertices f(x)=-2x^2-3x+1 is Maximum (-3/4 , 17/8)

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vertices f(x)=-2x^2-3x+1

vertices

Maximum (- \frac{3}{4}, \frac{17}{8})

Solution steps

y = - 2 x^{2} - 3 x + 1

The parabola parameters are:

a = - 2, b = - 3, c = 1

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

x_{v} = - \frac{( - 3 )}{2 ( - 2 )}

Simplify

x_{v} = - \frac{3}{4}

Plug in

x_{v} = - \frac{3}{4}

to find the

y_{v}

value

y_{v} = \frac{17}{8}

Therefore the parabola vertex is

(- \frac{3}{4}, \frac{17}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (- \frac{3}{4}, \frac{17}{8})

What is the vertices f(x)=-2x^2-3x+1 ?
The vertices f(x)=-2x^2-3x+1 is Maximum (-3/4 , 17/8)