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

The vertices y=x^2-2x+10 is Minimum (1,9)

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vertices y=x^2-2x+10

vertices

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

Minimum (1, 9)

Solution steps

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

The parabola parameters are:

a = 1, b = - 2, c = 10

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

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

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 9

Therefore the parabola vertex is

(1, 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, 9)

3 x - 2 y = 14

\frac{2}{x ^{2} + 1}

f (x) = x^{2} + 11 x

f (x) = \frac{1}{3} x - 1

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