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

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

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

vertices

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

Maximum (- 1, 3)

Solution steps

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

The parabola parameters are:

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

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

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

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(- 1, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 1, 3)

x^{2} + 4 y^{2} - 4 x = 0

4 x^{2} + 9 y^{2} = 36

3 x^{2} + 2 y^{2} + 6 x - 4 y + 5 = 18

f (x) = x^{2} + 8 x + 15

y^{2} = - 20 x

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