What is the vertices f(x)=2x^2-10x+12 ?

The vertices f(x)=2x^2-10x+12 is Minimum (5/2 ,-1/2)

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vertices f(x)=2x^2-10x+12

vertices

Minimum (\frac{5}{2}, - \frac{1}{2})

Solution steps

y = 2 x^{2} - 10 x + 12

The parabola parameters are:

a = 2, b = - 10, c = 12

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

x_{v} = - \frac{( - 10 )}{2 \cdot 2}

Simplify

- \frac{- 10}{2 \cdot 2} : \frac{5}{2}

x_{v} = \frac{5}{2}

Plug in

x_{v} = \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{1}{2}

Therefore the parabola vertex is

(\frac{5}{2}, - \frac{1}{2})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (\frac{5}{2}, - \frac{1}{2})

What is the vertices f(x)=2x^2-10x+12 ?
The vertices f(x)=2x^2-10x+12 is Minimum (5/2 ,-1/2)