What is the vertices f(x)=x^2+6x+4 ?

The vertices f(x)=x^2+6x+4 is Minimum (-3,-5)

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vertices f(x)=x^2+6x+4

vertices

f (x) = x^{2} + 6 x + 4

Minimum (- 3, - 5)

Solution steps

y = x^{2} + 6 x + 4

The parabola parameters are:

a = 1, b = 6, c = 4

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} = - 5

Therefore the parabola vertex is

(- 3, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, - 5)

f (x) = 2 x^{2} - 10 x + 12

\frac{x ^{3} - 1}{x - 1}

+ 10 x^{2} - 15 x + 2

f (x) = 4 x^{2} + 8 x - 8

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

What is the vertices f(x)=x^2+6x+4 ?
The vertices f(x)=x^2+6x+4 is Minimum (-3,-5)