What is the vertices y=2x^2-3x+1 ?

The vertices y=2x^2-3x+1 is Minimum (3/4 ,-1/8)

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vertices y=2x^2-3x+1

vertices

y = 2 x^{2} - 3 x + 1

Minimum (\frac{3}{4}, - \frac{1}{8})

Solution steps

y = 2 x^{2} - 3 x + 1

The parabola parameters are:

a = 2, b = - 3, c = 1

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

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

Simplify

x_{v} = \frac{3}{4}

Plug in

x_{v} = \frac{3}{4}

to find the

y_{v}

value

y_{v} = - \frac{1}{8}

Therefore the parabola vertex is

(\frac{3}{4}, - \frac{1}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (\frac{3}{4}, - \frac{1}{8})

y^{2} - 2 y - 12 x + 25 = 0

x^{2} - 4 x - 6

y = 84 x - x^{2}

\frac{x ^{2} x ^{1}}{x}

y = x^{2} - 2 x

What is the vertices y=2x^2-3x+1 ?
The vertices y=2x^2-3x+1 is Minimum (3/4 ,-1/8)