What is the vertices 3x^2+19x-40 ?

The vertices 3x^2+19x-40 is Minimum (-19/6 ,-841/12)

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vertices 3x^2+19x-40

vertices

3 x^{2} + 19 x - 40

Minimum (- \frac{19}{6}, - \frac{841}{12})

Solution steps

y = 3 x^{2} + 19 x - 40

The parabola parameters are:

a = 3, b = 19, c = - 40

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

x_{v} = - \frac{19}{2 \cdot 3}

Simplify

x_{v} = - \frac{19}{6}

Plug in

x_{v} = - \frac{19}{6}

to find the

y_{v}

value

y_{v} = - \frac{841}{12}

Therefore the parabola vertex is

(- \frac{19}{6}, - \frac{841}{12})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- \frac{19}{6}, - \frac{841}{12})

(x + 4)^{2} = 12 y - 24

x^{2} - 10 x - 4 y + 21 = 0

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

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

f (x) = 108 - 5 x^{2} - 7 x

What is the vertices 3x^2+19x-40 ?
The vertices 3x^2+19x-40 is Minimum (-19/6 ,-841/12)