What is the vertices f(x)=x^2+40x+20 ?

The vertices f(x)=x^2+40x+20 is Minimum (-20,-380)

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vertices f(x)=x^2+40x+20

vertices

f (x) = x^{2} + 40 x + 20

Minimum (- 20, - 380)

Solution steps

y = x^{2} + 40 x + 20

The parabola parameters are:

a = 1, b = 40, c = 20

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

x_{v} = - \frac{40}{2 \cdot 1}

Simplify

x_{v} = - 20

Plug in

x_{v} = - 20

to find the

y_{v}

value

y_{v} = - 380

Therefore the parabola vertex is

(- 20, - 380)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 20, - 380)

f (x) = - x^{2} + 4 x + 7

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

f (x) = x^{2} - 12 x + 36

y = x^{2} + 4 x + 1

f (x) = x^{2} + 16 x + 63

What is the vertices f(x)=x^2+40x+20 ?
The vertices f(x)=x^2+40x+20 is Minimum (-20,-380)