What is the vertices (x-5)^2=4(y-3) ?

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

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vertices (x-5)^2=4(y-3)

vertices

(x - 5)^{2} = 4 (y - 3)

Minimum (5, 3)

Solution steps

(x - 5)^{2} = 4 (y - 3)

Rewrite

(x - 5)^{2} = 4 (y - 3)

in the form

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

y = \frac{1 \cdot x ^{2}}{4} - \frac{5 x}{2} + \frac{25}{4} + 3

The parabola parameters are:

a = \frac{1}{4}, b = - \frac{5}{2}, c = \frac{25}{4} + 3

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

x_{v} = - \frac{( - \frac{5}{2} )}{2 ( \frac{1}{4} )}

Simplify

- \frac{- \frac{5}{2}}{2 ( \frac{1}{4} )} : 5

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(5, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{4}

Minimum (5, 3)

(x - 4)^{2} = 8 y + 16

- 4 x^{2}

x^{2} - 4 x - 7

f (x) = - 3 x^{2} + 4 x - 1

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

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