What is the vertices y=-x^2+4x+1 ?

The vertices y=-x^2+4x+1 is Maximum (2,5)

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vertices y=-x^2+4x+1

vertices

y = - x^{2} + 4 x + 1

Maximum (2, 5)

Solution steps

y = - x^{2} + 4 x + 1

The parabola parameters are:

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

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} = 5

Therefore the parabola vertex is

(2, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (2, 5)

f (x) = - x^{2} + 9

(x - 5)^{2}

f (x) = x + e^{- 3 x}, - 2 \leq x \leq 2

f (x) = - (x + 1) (x - 2) (x - 3)