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

The vertices f(x)=-x^2+2 is Maximum (0,2)

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

vertices

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

Maximum (0, 2)

Solution steps

y = - x^{2} + 2

The parabola parameters are:

a = - 1, b = 0, c = 2

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

x_{v} = - \frac{0}{2 ( - 1 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(0, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (0, 2)

f (x) = x^{2} + 5 x + 4

+ y^{2} + 12 x + 4 y - 32 = 0

f (x) = - 2 x^{2} - 16 x - 30

x = - 5 y^{2}

y = 6 x^{2} - 15 x