What is the vertices y=x^2+10x+25 ?

The vertices y=x^2+10x+25 is Minimum (-5,0)

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vertices y=x^2+10x+25

vertices

y = x^{2} + 10 x + 25

Minimum (- 5, 0)

Solution steps

y = x^{2} + 10 x + 25

The parabola parameters are:

a = 1, b = 10, c = 25

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

x_{v} = - \frac{10}{2 \cdot 1}

Simplify

x_{v} = - 5

Plug in

x_{v} = - 5

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(- 5, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 5, 0)

\frac{1}{1 - \frac{1}{x - 2}}

f (x) = 3 sin (x - π) - 1

y = - 0.55 x^{2} + 0 x + 10

3 x - 2 y = - 12