What is the vertices y=(2x+5)^2 ?

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

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vertices y=(2x+5)^2

vertices

y = (2 x + 5)^{2}

Minimum (- \frac{5}{2}, 0)

Solution steps

y = (2 x + 5)^{2}

y = (2 x + 5)^{2}

The parabola parameters are:

a = 4, m = - \frac{5}{2}, n = - \frac{5}{2}

x_{v} = \frac{m + n}{2}

x_{v} = \frac{( - \frac{5}{2} ) + ( - \frac{5}{2} )}{2}

Simplify

\frac{( - \frac{5}{2} ) + ( - \frac{5}{2} )}{2} : - \frac{5}{2}

x_{v} = - \frac{5}{2}

Plug in

x_{v} = - \frac{5}{2}

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(- \frac{5}{2}, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (- \frac{5}{2}, 0)

f (x) = - 9 x^{2} + 9 x + 1

+ x^{2} + 4 x - 1

y^{2} - 12 y - 4 x = - 28

24000 x^{2} + 900 x - 150

y = x^{2} + 4 x - 8