What is the vertices x^2-6x-4y+5=0 ?

The vertices x^2-6x-4y+5=0 is Minimum (3,-1)

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vertices x^2-6x-4y+5=0

vertices

x^{2} - 6 x - 4 y + 5 = 0

Minimum (3, - 1)

Solution steps

x^{2} - 6 x - 4 y + 5 = 0

Rewrite

x^{2} - 6 x - 4 y + 5 = 0

in the form

y = a x^{2} + b x + c

y = \frac{1 \cdot x ^{2}}{4} - \frac{3 x}{2} + \frac{5}{4}

The parabola parameters are:

a = \frac{1}{4}, b = - \frac{3}{2}, c = \frac{5}{4}

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

x_{v} = - \frac{( - \frac{3}{2} )}{2 ( \frac{1}{4} )}

Simplify

- \frac{- \frac{3}{2}}{2 ( \frac{1}{4} )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(3, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{4}

Minimum (3, - 1)

(x + 1)^{2} + 1

y = x^{2} - 5 x - 6

- x^{2} - 2 x + 8

y = 4 x^{2} - 8

f (x) = 2 x^{2} - 12 x

What is the vertices x^2-6x-4y+5=0 ?
The vertices x^2-6x-4y+5=0 is Minimum (3,-1)