What is the vertices y=3x^2-12x+10 ?

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

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vertices y=3x^2-12x+10

vertices

y = 3 x^{2} - 12 x + 10

Minimum (2, - 2)

Solution steps

y = 3 x^{2} - 12 x + 10

The parabola parameters are:

a = 3, b = - 12, c = 10

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

x_{v} = - \frac{( - 12 )}{2 \cdot 3}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(2, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (2, - 2)

f (x) = - 2 x^{2} - 4 x + 13

y^{2} = - 16 x

y = x^{2} - 6 x - 1

y = x^{2} - 5 x - 24

2 x^{2} + 5 x - 12

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