What is the vertices y=(-3x^2-2x+6)/5 ?

The vertices y=(-3x^2-2x+6)/5 is Maximum (-1/3 , 19/15)

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vertices y=(-3x^2-2x+6)/5

vertices

y = \frac{- 3 x ^{2} - 2 x + 6}{5}

Maximum (- \frac{1}{3}, \frac{19}{15})

Solution steps

y = \frac{- 3 x ^{2} - 2 x + 6}{5}

Rewrite

y = \frac{- 3 x ^{2} - 2 x + 6}{5}

in the form

y = a x^{2} + b x + c

y = - \frac{3 x ^{2}}{5} - \frac{2 x}{5} + \frac{6}{5}

The parabola parameters are:

a = - \frac{3}{5}, b = - \frac{2}{5}, c = \frac{6}{5}

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

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

Simplify

- \frac{- \frac{2}{5}}{2 ( - \frac{3}{5} )} : - \frac{1}{3}

x_{v} = - \frac{1}{3}

Plug in

x_{v} = - \frac{1}{3}

to find the

y_{v}

value

y_{v} = \frac{19}{15}

Therefore the parabola vertex is

(- \frac{1}{3}, \frac{19}{15})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{3}{5}

Maximum (- \frac{1}{3}, \frac{19}{15})

(x + 5)^{2}

y = 4 x^{2} - 4 x^{1}

y = - x^{2} + 2 x - 10

1 - x - x^{2}

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

What is the vertices y=(-3x^2-2x+6)/5 ?
The vertices y=(-3x^2-2x+6)/5 is Maximum (-1/3 , 19/15)