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

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

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vertices y=x^2-6x

vertices

y = 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)

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1

y^{2} + 12 x - 2 y - 11 = 0

x^{2} + 3 y^{2} = 1

x^{2} = 16 y