What is the vertices x^2-2x-4y+13=0 ?

The vertices x^2-2x-4y+13=0 is Minimum (1,3)

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vertices x^2-2x-4y+13=0

vertices

x^{2} - 2 x - 4 y + 13 = 0

Minimum (1, 3)

Solution steps

x^{2} - 2 x - 4 y + 13 = 0

Rewrite

x^{2} - 2 x - 4 y + 13 = 0

in the form

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

y = \frac{1 \cdot x ^{2}}{4} - \frac{x}{2} + \frac{13}{4}

The parabola parameters are:

a = \frac{1}{4}, b = - \frac{1}{2}, c = \frac{13}{4}

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

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

Simplify

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

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(1, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{4}

Minimum (1, 3)

5 x^{2} - 10

f (x) = 2 x^{2} - 4 x + 3

y = 18 x^{2} - 6 x + 4

f (x) = - x^{2} + 14 x - 48

f (x) = x^{2} + 11 x - 35

What is the vertices x^2-2x-4y+13=0 ?
The vertices x^2-2x-4y+13=0 is Minimum (1,3)