What is the vertices f(x)=x^2+9x+8 ?

The vertices f(x)=x^2+9x+8 is Minimum (-9/2 ,-49/4)

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vertices f(x)=x^2+9x+8

vertices

f (x) = x^{2} + 9 x + 8

Minimum (- \frac{9}{2}, - \frac{49}{4})

Solution steps

y = x^{2} + 9 x + 8

The parabola parameters are:

a = 1, b = 9, c = 8

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

x_{v} = - \frac{9}{2 \cdot 1}

Simplify

x_{v} = - \frac{9}{2}

Plug in

x_{v} = - \frac{9}{2}

to find the

y_{v}

value

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

Therefore the parabola vertex is

(- \frac{9}{2}, - \frac{49}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{9}{2}, - \frac{49}{4})

8 x^{2} + 12 x + 3

- 5 x^{2} + 118 x + 7

y^{2} = - 6 x

4 x - 2 x^{2}

y = \frac{95 x ^{2} - 10 x + 26}{10}

What is the vertices f(x)=x^2+9x+8 ?
The vertices f(x)=x^2+9x+8 is Minimum (-9/2 ,-49/4)