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

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

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

vertices

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

Minimum (2, 5)

Solution steps

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

Rewrite

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

in the form

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

y = 3 x^{2} - 12 x + 17

The parabola parameters are:

a = 3, b = - 12, c = 17

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

x_{v} = - \frac{( - 12 )}{2 \cdot 3}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(2, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (2, 5)

y = 2.6 + 2.2 x - 0.6 x^{2}

f (x) = (x - 1)^{2}

V (x) = 325 x^{2} - 4600 x + 145000

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

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

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