What is the vertices y=3x^2-24x-15 ?

The vertices y=3x^2-24x-15 is Minimum (4,-63)

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vertices y=3x^2-24x-15

vertices

y = 3 x^{2} - 24 x - 15

Minimum (4, - 63)

Solution steps

y = 3 x^{2} - 24 x - 15

The parabola parameters are:

a = 3, b = - 24, c = - 15

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

x_{v} = - \frac{( - 24 )}{2 \cdot 3}

Simplify

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = - 63

Therefore the parabola vertex is

(4, - 63)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (4, - 63)

f (x) = 3 x^{2} - 4 x - 4

y = - x^{2} + 6 x - 14

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

\frac{7 x}{9 x - 1}

1 + x - 2

What is the vertices y=3x^2-24x-15 ?
The vertices y=3x^2-24x-15 is Minimum (4,-63)