What is the vertices f(x)=x^2-6x+10 ?

The vertices f(x)=x^2-6x+10 is Minimum (3,1)

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vertices f(x)=x^2-6x+10

vertices

f (x) = x^{2} - 6 x + 10

Minimum (3, 1)

Solution steps

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

The parabola parameters are:

a = 1, b = - 6, c = 10

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

x_{v} = - \frac{( - 6 )}{2 \cdot 1}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(3, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, 1)

- x^{2} + 4 x

f (x) = - 6 x^{2} + 17 x - 12

f (x) = 4 x^{2} - 16 x + 10

2 x^{2} - 6 x + 4

f (x) = - 3 x^{2} - 18 x + 2

What is the vertices f(x)=x^2-6x+10 ?
The vertices f(x)=x^2-6x+10 is Minimum (3,1)