What is the vertices f(x)=x^2+8x+15 ?

The vertices f(x)=x^2+8x+15 is Minimum (-4,-1)

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vertices f(x)=x^2+8x+15

vertices

f (x) = x^{2} + 8 x + 15

Minimum (- 4, - 1)

Solution steps

y = x^{2} + 8 x + 15

The parabola parameters are:

a = 1, b = 8, c = 15

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

x_{v} = - \frac{8}{2 \cdot 1}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(- 4, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, - 1)

y^{2} = - 20 x

f (x) = - \frac{x ^{2}}{10} + \frac{9 x}{10} + \frac{11}{5}

y^{2} = 32 x

y + 4 x \leq x^{2} + 2

y^{2} - y - x + 1 = 0

What is the vertices f(x)=x^2+8x+15 ?
The vertices f(x)=x^2+8x+15 is Minimum (-4,-1)