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

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

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vertices y=x^2-4x

vertices

y = x^{2} - 4 x

Minimum (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 \cdot 1}

Simplify

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

Minimum (2, - 4)

x sin (\frac{1}{x})

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

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

f (x) = \frac{9 e ^{x}}{e ^{x} - 2}

f (x) = - 2 cos (4 π x)