What is the vertices y=x^2+6x+7 ?

The vertices y=x^2+6x+7 is Minimum (-3,-2)

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vertices y=x^2+6x+7

vertices

y = x^{2} + 6 x + 7

Minimum (- 3, - 2)

Solution steps

y = x^{2} + 6 x + 7

The parabola parameters are:

a = 1, b = 6, c = 7

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

x_{v} = - \frac{6}{2 \cdot 1}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(- 3, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, - 2)

f (x) = \frac{x + 2}{3 x - 15}

f (x) = \frac{x ^{2} + 1}{2 x ^{2} - 3 x - 2}

3 x + 2 y = 7

f (x) = 4 x^{3} - 48 x - 8

f (x) = 7 x + 6 y = 6