What is the vertices-3x^2+5x ?

The vertices-3x^2+5x is Maximum (5/6 , 25/12)

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vertices-3x^2+5x

vertices

- 3 x^{2} + 5 x

Maximum (\frac{5}{6}, \frac{25}{12})

Solution steps

y = - 3 x^{2} + 5 x

The parabola parameters are:

a = - 3, b = 5, c = 0

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

x_{v} = - \frac{5}{2 ( - 3 )}

Simplify

x_{v} = \frac{5}{6}

Plug in

x_{v} = \frac{5}{6}

to find the

y_{v}

value

y_{v} = \frac{25}{12}

Therefore the parabola vertex is

(\frac{5}{6}, \frac{25}{12})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (\frac{5}{6}, \frac{25}{12})

y = 4 x + 4 x - 2 + x^{2} - 3

y = 3 - x^{2}

y = 25 x^{2} - 36

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

y^{2} = 4 x