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

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

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vertices y=-x^2+4x

vertices

y = - x^{2} + 4 x

Maximum (2, 4)

Solution steps

y = - x^{2} + 4 x

The parabola parameters are:

a = - 1, b = 4, c = 0

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} = 4

Therefore the parabola vertex is

(2, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (2, 4)

y = 4^{x} + 8

f (x) = x^{2} - 4 x + 3

\frac{1}{x - 4}

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

f (x) = \frac{x ^{2} + 1}{x + 1}