What is the vertices y=-x^2+1 ?

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

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vertices y=-x^2+1

vertices

y = - 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)

f (x) = - 2 x^{2} + 20 x + 48

y = x^{2} - 15 x

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

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

f (x) = - 0.02 x^{2} + x + 6