What is the vertices y=6x^2-15x ?

The vertices y=6x^2-15x is Minimum (5/4 ,-75/8)

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vertices y=6x^2-15x

vertices

y = 6 x^{2} - 15 x

Minimum (\frac{5}{4}, - \frac{75}{8})

Solution steps

y = 6 x^{2} - 15 x

The parabola parameters are:

a = 6, b = - 15, c = 0

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

x_{v} = - \frac{( - 15 )}{2 \cdot 6}

Simplify

- \frac{- 15}{2 \cdot 6} : \frac{5}{4}

x_{v} = \frac{5}{4}

Plug in

x_{v} = \frac{5}{4}

to find the

y_{v}

value

y_{v} = - \frac{75}{8}

Therefore the parabola vertex is

(\frac{5}{4}, - \frac{75}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 6

Minimum (\frac{5}{4}, - \frac{75}{8})

f (x) = - (x - 5)^{2} - 4

y = \frac{x ^{2} 5 x 8}{x 3}

x^{2} - 4 x

t^{2} + 2 t - 48

f (x) = - x^{2} + 2 x + 8

What is the vertices y=6x^2-15x ?
The vertices y=6x^2-15x is Minimum (5/4 ,-75/8)