What is the vertices y=x^2+8x+3 ?

The vertices y=x^2+8x+3 is Minimum (-4,-13)

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vertices y=x^2+8x+3

vertices

y = x^{2} + 8 x + 3

Minimum (- 4, - 13)

Solution steps

y = x^{2} + 8 x + 3

The parabola parameters are:

a = 1, b = 8, c = 3

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} = - 13

Therefore the parabola vertex is

(- 4, - 13)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, - 13)

f (x) = (x + 4.4)^{20} - 1.2

f (x) = \frac{5}{x - 6}

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

3 x

y = - \frac{3}{4} x + 1, (9, 12)