What is the vertices y=-x^2+5 ?

The vertices y=-x^2+5 is Maximum (0,5)

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vertices y=-x^2+5

vertices

y = - x^{2} + 5

Maximum (0, 5)

Solution steps

y = - x^{2} + 5

The parabola parameters are:

a = - 1, b = 0, c = 5

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} = 5

Therefore the parabola vertex is

(0, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (0, 5)

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

f (x) = 3 x^{2} - 2 x - 1

7 x^{2} - 40 x - 25

f (x) = (x - 3)^{2} - 1

f (x) = x^{2} + 6 x + 7