What is the vertices y=-x^2+4x+5 ?

The vertices y=-x^2+4x+5 is Maximum (2,9)

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vertices y=-x^2+4x+5

vertices

y = - x^{2} + 4 x + 5

Maximum (2, 9)

Solution steps

y = - x^{2} + 4 x + 5

The parabola parameters are:

a = - 1, b = 4, c = 5

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

x_{v} = - \frac{4}{2 ( - 1 )}

Simplify

- \frac{4}{2 ( - 1 )} : 2

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 9

Therefore the parabola vertex is

(2, 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (2, 9)

y = 2 x^{2} + 15 x

f (x) = - x^{2} + 10 x - 16

f (x) = - x^{2} + 5 x - 6

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

y = 3 x^{2} - 3 x - 6