What is the vertices (5x^2)/2 ?

The vertices (5x^2)/2 is Minimum (0,0)

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vertices (5x^2)/2

vertices

\frac{5 x ^{2}}{2}

Minimum (0, 0)

Solution steps

y = \frac{5 x ^{2}}{2}

The parabola parameters are:

a = \frac{5}{2}, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( \frac{5}{2} )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(0, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{5}{2}

Minimum (0, 0)

x^{2} - 2 x

- x^{2} - 2 x + 3

y^{2} = 6 x - 32

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

2 x^{2} - 4 x