What is the vertices f(x)=x^2+4x+1 ?

The vertices f(x)=x^2+4x+1 is Minimum (-2,-3)

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vertices f(x)=x^2+4x+1

vertices

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

Minimum (- 2, - 3)

Solution steps

y = x^{2} + 4 x + 1

The parabola parameters are:

a = 1, b = 4, c = 1

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

x_{v} = - \frac{4}{2 \cdot 1}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 3

Therefore the parabola vertex is

(- 2, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 3)

\frac{x ^{2}}{9} - \frac{y ^{2}}{25} = 1

x - 2 = \frac{1}{8} (y + 1)^{2}

y = 2 x^{2}

\frac{y ^{2}}{9} - \frac{x ^{2}}{16} = 1

2 x^{2} + 7 y^{2} = 14

What is the vertices f(x)=x^2+4x+1 ?
The vertices f(x)=x^2+4x+1 is Minimum (-2,-3)