What is the vertices y=x^2+4x+1 ?

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

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vertices y=x^2+4x+1

vertices

y = 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)

f (x) = x^{2} + 16 x + 63

y = 2 x^{2} - 4 x + 7

y = (x + 3) (x - 3)

f (x) = - 3 x^{2} - 18 x - 25

y = 18 x^{2} - 14 x + 9