What is the vertices y=x^2-6x+18 ?

The vertices y=x^2-6x+18 is Minimum (3,9)

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vertices y=x^2-6x+18

vertices

y = x^{2} - 6 x + 18

Minimum (3, 9)

Solution steps

y = x^{2} - 6 x + 18

The parabola parameters are:

a = 1, b = - 6, c = 18

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} = 9

Therefore the parabola vertex is

(3, 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, 9)

(x + 6)^{5}

\frac{x + 7}{x - 1}

f (x) = x^{4} + 1

f (x) = (x + 3)^{\frac{1}{3}} - 4

f (x) = \frac{6 x}{x + 8}