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

The vertices 3x^2-5x+2 is Minimum (5/6 ,-1/12)

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vertices 3x^2-5x+2

vertices

3 x^{2} - 5 x + 2

Minimum (\frac{5}{6}, - \frac{1}{12})

Solution steps

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

The parabola parameters are:

a = 3, b = - 5, c = 2

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{1}{12}

Therefore the parabola vertex is

(\frac{5}{6}, - \frac{1}{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{1}{12})

y = - 3 x^{2} + 12 x - 8

x^{2} + 8 x + 18

x = - 2 (y + 3)^{2} + 4

x^{2} + 5 x - 14

f (x) = x^{2} + 8 x + 7