What is the vertices y=x^2-6x+7 ?

The vertices y=x^2-6x+7 is Minimum (3,-2)

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vertices y=x^2-6x+7

vertices

y = x^{2} - 6 x + 7

Minimum (3, - 2)

Solution steps

y = x^{2} - 6 x + 7

The parabola parameters are:

a = 1, b = - 6, c = 7

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} = - 2

Therefore the parabola vertex is

(3, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, - 2)

\frac{x ^{3}}{x ^{2} - 1}

f (x) = ∣ x - 2 ∣

x^{3} - 4 x^{2} + 8 x - 5

6 x^{4} + 8 x^{3}

\frac{1}{x}