What is the vertices y=x^2+2x+6 ?

The vertices y=x^2+2x+6 is Minimum (-1,5)

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vertices y=x^2+2x+6

vertices

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

Minimum (- 1, 5)

Solution steps

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

The parabola parameters are:

a = 1, b = 2, c = 6

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

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

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(- 1, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, 5)

x + 1.2 y = 12.4

f (x) = e^{\frac{x - 1}{x ^{2} - 1}}

f (x) = 9 - x^{2}

f (x) = x^{4} + 5 x^{3} - 9 x^{2} - 45 x

f (x) = \frac{6 + x}{6 - x}