What is the vertices y=x^2+2x+18 ?

The vertices y=x^2+2x+18 is Minimum (-1,17)

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vertices y=x^2+2x+18

vertices

y = x^{2} + 2 x + 18

Minimum (- 1, 17)

Solution steps

y = x^{2} + 2 x + 18

The parabola parameters are:

a = 1, b = 2, c = 18

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} = 17

Therefore the parabola vertex is

(- 1, 17)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, 17)

f (x) = x^{2} - 6 x + 8

\frac{2 x}{x ^{2} + 2 x}

(4, 0), (20, 18)

(3 - x) e^{- x}

y = \frac{1}{x}