What is the vertices y=x(2x+3) ?

The vertices y=x(2x+3) is Minimum (-3/4 ,-9/8)

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vertices y=x(2x+3)

vertices

y = x (2 x + 3)

Minimum (- \frac{3}{4}, - \frac{9}{8})

Solution steps

y = x (2 x + 3)

y = x (2 x + 3)

The parabola parameters are:

a = 2, m = 0, n = - \frac{3}{2}

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

x_{v} = \frac{0 + ( - \frac{3}{2} )}{2}

Simplify

\frac{0 + ( - \frac{3}{2} )}{2} : - \frac{3}{4}

x_{v} = - \frac{3}{4}

Plug in

x_{v} = - \frac{3}{4}

to find the

y_{v}

value

y_{v} = - \frac{9}{8}

Therefore the parabola vertex is

(- \frac{3}{4}, - \frac{9}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- \frac{3}{4}, - \frac{9}{8})

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

f (x) = x^{2} - 4 x + 5

y^{2} = 8 x

x \leq y^{2}

\frac{1}{2} x^{2} + 8 x + 5