What is the vertices y=2-x-2x^2 ?

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

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vertices y=2-x-2x^2

vertices

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

Maximum (- \frac{1}{4}, \frac{17}{8})

Solution steps

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

Rewrite

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

in the form

y = a x^{2} + b x + c

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

The parabola parameters are:

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

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{17}{8}

Therefore the parabola vertex is

(- \frac{1}{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{1}{4}, \frac{17}{8})

\frac{x ^{2}}{36} + \frac{y ^{2}}{9} = 1

x^{2} + y^{2} - 14 x + 2 y + 41 = 0

y = x^{2} + 3

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1

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