What is the vertices y=x^2+2x+5 ?

The vertices y=x^2+2x+5 is Minimum (-1,4)

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vertices y=x^2+2x+5

vertices

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

Minimum (- 1, 4)

Solution steps

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

The parabola parameters are:

a = 1, b = 2, c = 5

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} = 4

Therefore the parabola vertex is

(- 1, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, 4)

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

f (x) = x^{2} - 12 x

f (x) = x^{2} + 4 x - 21

5 + 4 x - x^{2}

8 - 5 x^{2}