What is the vertices f(x)=3(x+2)^2+1 ?

The vertices f(x)=3(x+2)^2+1 is Minimum (-2,1)

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vertices f(x)=3(x+2)^2+1

vertices

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

Minimum (- 2, 1)

Solution steps

y = 3 (x + 2)^{2} + 1

Rewrite

y = 3 (x + 2)^{2} + 1

in the form

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

y = 3 x^{2} + 12 x + 13

The parabola parameters are:

a = 3, b = 12, c = 13

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} = 1

Therefore the parabola vertex is

(- 2, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- 2, 1)

- 2 x^{2} + 8 x - 1

y = x^{2} + 3 x

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

x = - 3 y^{2} + 5

f (x) = x^{2} + 10 x + 6

What is the vertices f(x)=3(x+2)^2+1 ?
The vertices f(x)=3(x+2)^2+1 is Minimum (-2,1)