What is the vertices 3x^2+5x-12 ?

The vertices 3x^2+5x-12 is Minimum (-5/6 ,-169/12)

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vertices 3x^2+5x-12

vertices

3 x^{2} + 5 x - 12

Minimum (- \frac{5}{6}, - \frac{169}{12})

Solution steps

y = 3 x^{2} + 5 x - 12

The parabola parameters are:

a = 3, b = 5, c = - 12

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{169}{12}

Therefore the parabola vertex is

(- \frac{5}{6}, - \frac{169}{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{169}{12})

(x - 1) (x - 3)

y = x^{2} - 8 x + 15

y^{2} = - 28 x

x^{2} - 4 x - 5

x^{2} - 7

What is the vertices 3x^2+5x-12 ?
The vertices 3x^2+5x-12 is Minimum (-5/6 ,-169/12)