What is the vertices f(x)=x^2+6x+8 ?

The vertices f(x)=x^2+6x+8 is Minimum (-3,-1)

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vertices f(x)=x^2+6x+8

vertices

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

Minimum (- 3, - 1)

Solution steps

y = x^{2} + 6 x + 8

The parabola parameters are:

a = 1, b = 6, c = 8

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} = - 1

Therefore the parabola vertex is

(- 3, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, - 1)

4 x^{2} + 16 y^{2} - 4 x - 32 y + 1 = 0

x = (y + 2)^{2}

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

f (x) = x^{2} + 5 x + 6

2 x^{2} + 2 y^{2} - 2 x - 2 y - 3 = 0

What is the vertices f(x)=x^2+6x+8 ?
The vertices f(x)=x^2+6x+8 is Minimum (-3,-1)