What is the vertices y=3x^2+5x-4 ?

The vertices y=3x^2+5x-4 is Minimum (-5/6 ,-73/12)

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vertices y=3x^2+5x-4

vertices

y = 3 x^{2} + 5 x - 4

Minimum (- \frac{5}{6}, - \frac{73}{12})

Solution steps

y = 3 x^{2} + 5 x - 4

The parabola parameters are:

a = 3, b = 5, c = - 4

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

x_{v} = - \frac{5}{2 \cdot 3}

Simplify

x_{v} = - \frac{5}{6}

Plug in

x_{v} = - \frac{5}{6}

to find the

y_{v}

value

y_{v} = - \frac{73}{12}

Therefore the parabola vertex is

(- \frac{5}{6}, - \frac{73}{12})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- \frac{5}{6}, - \frac{73}{12})

y = x^{2} + 4

x^{2} - 6 x + 5

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

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

- x^{2} + 6 x - 5

What is the vertices y=3x^2+5x-4 ?
The vertices y=3x^2+5x-4 is Minimum (-5/6 ,-73/12)