What is the vertices y=24000x^2+900x-150 ?

The vertices y=24000x^2+900x-150 is Minimum (-3/160 ,-2535/16)

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vertices y=24000x^2+900x-150

vertices

y = 24000 x^{2} + 900 x - 150

Minimum (- \frac{3}{160}, - \frac{2535}{16})

Solution steps

y = 24000 x^{2} + 900 x - 150

The parabola parameters are:

a = 24000, b = 900, c = - 150

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

x_{v} = - \frac{900}{2 \cdot 24000}

Simplify

- \frac{900}{2 \cdot 24000} : - \frac{3}{160}

x_{v} = - \frac{3}{160}

Plug in

x_{v} = - \frac{3}{160}

to find the

y_{v}

value

y_{v} = - \frac{2535}{16}

Therefore the parabola vertex is

(- \frac{3}{160}, - \frac{2535}{16})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 24000

Minimum (- \frac{3}{160}, - \frac{2535}{16})

16 - (12 x + 8)^{2}

y^{2} - 8 x - 6 y - 23 = 0

g (x) = (x + 3)^{2} + 5

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

f (x) = x^{2} + 8 + 15

What is the vertices y=24000x^2+900x-150 ?
The vertices y=24000x^2+900x-150 is Minimum (-3/160 ,-2535/16)