What is the vertices y=+x^2+8x+9 ?

The vertices y=+x^2+8x+9 is Minimum (-4,-7)

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vertices y=+x^2+8x+9

vertices

y = + x^{2} + 8 x + 9

Minimum (- 4, - 7)

Solution steps

y = x^{2} + 8 x + 9

The parabola parameters are:

a = 1, b = 8, c = 9

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

x_{v} = - \frac{8}{2 \cdot 1}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = - 7

Therefore the parabola vertex is

(- 4, - 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, - 7)

25 x^{2} - 36

y = 4 x^{2} - 40 x + 98

x = 2 y^{2}

x^{2} - 14 x + 49

y = x^{2} - 12 x + 36