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

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

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

vertices

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

Minimum (- 3, - 2)

Solution steps

y = \frac{1}{3} x^{2} + 2 x + 1

Rewrite

y = \frac{1}{3} x^{2} + 2 x + 1

in the form

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

y = \frac{x ^{2}}{3} + 2 x + 1

The parabola parameters are:

a = \frac{1}{3}, b = 2, c = 1

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

x_{v} = - \frac{2}{2 ( \frac{1}{3} )}

Simplify

- \frac{2}{2 ( \frac{1}{3} )} : - 3

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(- 3, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{3}

Minimum (- 3, - 2)

4 x^{2} - 8 x + 2 y = 5

y = 2 x^{2} + 5 x - 7

y = x (x - 18)

y = - x^{2} + 4

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

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