What is the vertices y=x^2+4x ?

The vertices y=x^2+4x is Minimum (-2,-4)

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vertices y=x^2+4x

vertices

y = x^{2} + 4 x

Minimum (- 2, - 4)

Solution steps

y = x^{2} + 4 x

The parabola parameters are:

a = 1, b = 4, c = 0

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

x_{v} = - \frac{4}{2 \cdot 1}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(- 2, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 4)

cot (x)

y = x + 9

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

f (x) = \frac{( x + 2 )}{( 3 x - 1 )}

3 x - y = - 1