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

The vertices 3x^2+2x-3 is Minimum (-1/3 ,-10/3)

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vertices 3x^2+2x-3

vertices

3 x^{2} + 2 x - 3

Minimum (- \frac{1}{3}, - \frac{10}{3})

Solution steps

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

The parabola parameters are:

a = 3, b = 2, c = - 3

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

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

Simplify

- \frac{2}{2 \cdot 3} : - \frac{1}{3}

x_{v} = - \frac{1}{3}

Plug in

x_{v} = - \frac{1}{3}

to find the

y_{v}

value

y_{v} = - \frac{10}{3}

Therefore the parabola vertex is

(- \frac{1}{3}, - \frac{10}{3})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- \frac{1}{3}, - \frac{10}{3})

2 x^{2} + 4 x + 1

- x^{2} + 2 x + 8

y = x^{2} + 2 x - 1

y = - 2 x^{2} - 4 x + 6

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

What is the vertices 3x^2+2x-3 ?
The vertices 3x^2+2x-3 is Minimum (-1/3 ,-10/3)