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

The vertices f(x)=-x^2+1 is Maximum (0,1)

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

vertices

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

Maximum (0, 1)

Solution steps

y = - x^{2} + 1

The parabola parameters are:

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

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} = 1

Therefore the parabola vertex is

(0, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (0, 1)

(x - 1)^{2} = 16 (y - 2)

- 0.5 x^{2} - 3

y^{2} - 8 x + 2 y + 17 = 0

y = x^{2} + 4 x + 3

y = x^{2} - 10 x + 7