What is the vertices f(x)=x^2+12x+35 ?

The vertices f(x)=x^2+12x+35 is Minimum (-6,-1)

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vertices f(x)=x^2+12x+35

vertices

f (x) = x^{2} + 12 x + 35

Minimum (- 6, - 1)

Solution steps

y = x^{2} + 12 x + 35

The parabola parameters are:

a = 1, b = 12, c = 35

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

x_{v} = - \frac{12}{2 \cdot 1}

Simplify

x_{v} = - 6

Plug in

x_{v} = - 6

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(- 6, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 6, - 1)

y^{2} + 14 y + 4 x + 45 = 0

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

x^{2} + 8 x + 12

x^{2} + 8 x + 13

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

What is the vertices f(x)=x^2+12x+35 ?
The vertices f(x)=x^2+12x+35 is Minimum (-6,-1)