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

The vertices f(x)=2x^2-12x+10 is Minimum (3,-8)

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

vertices

f (x) = 2 x^{2} - 12 x + 10

Minimum (3, - 8)

Solution steps

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

The parabola parameters are:

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

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

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

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 8

Therefore the parabola vertex is

(3, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (3, - 8)

18 x - 9 x^{2} - 9

+ s^{2} + 2 s + 2

f (x) = x^{2} + 7 x + 12

f (x) = x^{2} + 7 x + 10

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

What is the vertices f(x)=2x^2-12x+10 ?
The vertices f(x)=2x^2-12x+10 is Minimum (3,-8)