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

The vertices f(x)=x^2-6x is Minimum (3,-9)

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

vertices

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

Minimum (3, - 9)

Solution steps

y = x^{2} - 6 x

The parabola parameters are:

a = 1, b = - 6, c = 0

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)

y^{2} + 12 x + 4 y - 32 = 0

x^{2} + y^{2} - 18 x - 14 y + 124 = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = 1

9 (x - 1)^{2} - 16 (y + 2)^{2} = 144

9 x^{2} + 4 y^{2} + 36 x - 24 y + 36 = 0