What is the vertices f(x)=-1/3 x^2-7/6 x+1 ?

The vertices f(x)=-1/3 x^2-7/6 x+1 is Maximum (-7/4 , 49/48+1)

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vertices f(x)=-1/3 x^2-7/6 x+1

vertices

f (x) = - \frac{1}{3} x^{2} - \frac{7}{6} x + 1

Maximum (- \frac{7}{4}, \frac{49}{48} + 1)

Solution steps

y = - \frac{1}{3} x^{2} - \frac{7}{6} x + 1

Rewrite

y = - \frac{1}{3} x^{2} - \frac{7}{6} x + 1

in the form

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

y = - \frac{x ^{2}}{3} - \frac{7 x}{6} + 1

The parabola parameters are:

a = - \frac{1}{3}, b = - \frac{7}{6}, c = 1

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

x_{v} = - \frac{( - \frac{7}{6} )}{2 ( - \frac{1}{3} )}

Simplify

- \frac{- \frac{7}{6}}{2 ( - \frac{1}{3} )} : - \frac{7}{4}

x_{v} = - \frac{7}{4}

Plug in

x_{v} = - \frac{7}{4}

to find the

y_{v}

value

y_{v} = \frac{49}{48} + 1

Therefore the parabola vertex is

(- \frac{7}{4}, \frac{49}{48} + 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{3}

Maximum (- \frac{7}{4}, \frac{49}{48} + 1)

4 x^{2} + 4 x - 8

f (x) = - 2 x^{2} - 16 x - 31

- x^{2} + 2 x + 2

f (x) = - 2 x^{2} - 16 x - 37

- x^{2} + 2 x + 4

What is the vertices f(x)=-1/3 x^2-7/6 x+1 ?
The vertices f(x)=-1/3 x^2-7/6 x+1 is Maximum (-7/4 , 49/48+1)