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

The vertices y=x^2-5x is Minimum (5/2 ,-25/4)

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vertices y=x^2-5x

vertices

y = x^{2} - 5 x

Minimum (\frac{5}{2}, - \frac{25}{4})

Solution steps

y = x^{2} - 5 x

The parabola parameters are:

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

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

x_{v} = - \frac{( - 5 )}{2 \cdot 1}

Simplify

x_{v} = \frac{5}{2}

Plug in

x_{v} = \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{25}{4}

Therefore the parabola vertex is

(\frac{5}{2}, - \frac{25}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{5}{2}, - \frac{25}{4})

f (x) = - 36 - x^{2}

(1, 2), (3, 7)

(x - 2)^{3} + 1

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

(81) (8 - 7)