What is the vertices f(x)=-(x+3)(x+10) ?

The vertices f(x)=-(x+3)(x+10) is Maximum (-13/2 , 49/4)

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vertices f(x)=-(x+3)(x+10)

vertices

f (x) = - (x + 3) (x + 10)

Maximum (- \frac{13}{2}, \frac{49}{4})

Solution steps

y = - (x + 3) (x + 10)

y = - (x + 3) (x + 10)

The parabola parameters are:

a = - 1, m = - 3, n = - 10

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

x_{v} = \frac{( - 3 ) + ( - 10 )}{2}

Simplify

x_{v} = - \frac{13}{2}

Plug in

x_{v} = - \frac{13}{2}

to find the

y_{v}

value

y_{v} = \frac{49}{4}

Therefore the parabola vertex is

(- \frac{13}{2}, \frac{49}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- \frac{13}{2}, \frac{49}{4})

y^{2} = - 32 x

y = 2 x^{2} + 16 x + 29

f (x) = - (x^{2} + 3 x - 4)

y = 4 x^{2} - 16

f (x) = x^{2} - 8 x + 17

What is the vertices f(x)=-(x+3)(x+10) ?
The vertices f(x)=-(x+3)(x+10) is Maximum (-13/2 , 49/4)