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

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

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

vertices

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

Minimum (5, 2)

Solution steps

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

Rewrite

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

in the form

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

y = 3 x^{2} - 30 x + 77

The parabola parameters are:

a = 3, b = - 30, c = 77

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

x_{v} = - \frac{( - 30 )}{2 \cdot 3}

Simplify

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(5, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (5, 2)

y = (x + 1) (x + 2)

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

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

x^{2} - 8 x + 12

X^{2} - 5 X - 24

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