What is the vertices y=x^2-15x ?

The vertices y=x^2-15x is Minimum (15/2 ,-225/4)

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vertices y=x^2-15x

vertices

y = x^{2} - 15 x

Minimum (\frac{15}{2}, - \frac{225}{4})

Solution steps

y = x^{2} - 15 x

The parabola parameters are:

a = 1, b = - 15, c = 0

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

x_{v} = - \frac{( - 15 )}{2 \cdot 1}

Simplify

x_{v} = \frac{15}{2}

Plug in

x_{v} = \frac{15}{2}

to find the

y_{v}

value

y_{v} = - \frac{225}{4}

Therefore the parabola vertex is

(\frac{15}{2}, - \frac{225}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{15}{2}, - \frac{225}{4})

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

f (x) = (x - 2)^{2} - 1

f (x) = - 0.02 x^{2} + x + 6

y = x^{2} - 1

+ \frac{- 1}{2} x^{2} + 18

What is the vertices y=x^2-15x ?
The vertices y=x^2-15x is Minimum (15/2 ,-225/4)