What is the vertices f(x)=x^2-5x ?

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

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vertices f(x)=x^2-5x

vertices

f (x) = 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})

3 x^{2} + 2 y = 26

y^{2} = - 2 x

x^{2} + y^{2} - 4 x + 10 y + 13 = 0

x^{2} + y^{2} + 2 x + 6 y + 2 = 0

\frac{y ^{2}}{9} - \frac{x ^{2}}{16} = 1

What is the vertices f(x)=x^2-5x ?
The vertices f(x)=x^2-5x is Minimum (5/2 ,-25/4)