What is the vertices y=4x^2-12x+10 ?

The vertices y=4x^2-12x+10 is Minimum (3/2 ,1)

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vertices y=4x^2-12x+10

vertices

y = 4 x^{2} - 12 x + 10

Minimum (\frac{3}{2}, 1)

Solution steps

y = 4 x^{2} - 12 x + 10

The parabola parameters are:

a = 4, b = - 12, c = 10

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

x_{v} = - \frac{( - 12 )}{2 \cdot 4}

Simplify

- \frac{- 12}{2 \cdot 4} : \frac{3}{2}

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

Plug in

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

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(\frac{3}{2}, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (\frac{3}{2}, 1)

x^{2} - 12 x + 16 y + 68 = 0

f (x) = 4 - x^{2}

f (x) = x^{2} - 4 x + 7

y = x^{2} - 6 x + 10

y = 3 x^{2} - 30 x + 77

What is the vertices y=4x^2-12x+10 ?
The vertices y=4x^2-12x+10 is Minimum (3/2 ,1)