What is the vertices (x-1)^2=16(y-2) ?

The vertices (x-1)^2=16(y-2) is Minimum (1,2)

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vertices (x-1)^2=16(y-2)

vertices

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

Minimum (1, 2)

Solution steps

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

Rewrite

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

in the form

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

y = \frac{1 \cdot x ^{2}}{16} - \frac{x}{8} + \frac{1}{16} + 2

The parabola parameters are:

a = \frac{1}{16}, b = - \frac{1}{8}, c = \frac{1}{16} + 2

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

x_{v} = - \frac{( - \frac{1}{8} )}{2 ( \frac{1}{16} )}

Simplify

- \frac{- \frac{1}{8}}{2 ( \frac{1}{16} )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(1, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{16}

Minimum (1, 2)

- 0.5 x^{2} - 3

y^{2} - 8 x + 2 y + 17 = 0

y = x^{2} + 4 x + 3

y = x^{2} - 10 x + 7

y = 92 x^{2}

What is the vertices (x-1)^2=16(y-2) ?
The vertices (x-1)^2=16(y-2) is Minimum (1,2)