What is the vertices y=x^2+16x+71 ?

The vertices y=x^2+16x+71 is Minimum (-8,7)

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vertices y=x^2+16x+71

vertices

y = x^{2} + 16 x + 71

Minimum (- 8, 7)

Solution steps

y = x^{2} + 16 x + 71

The parabola parameters are:

a = 1, b = 16, c = 71

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

x_{v} = - \frac{16}{2 \cdot 1}

Simplify

x_{v} = - 8

Plug in

x_{v} = - 8

to find the

y_{v}

value

y_{v} = 7

Therefore the parabola vertex is

(- 8, 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 8, 7)

f (x) = 2 x^{2} - 12 x + 1

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

8 x^{2} + 12 x + 3

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

y^{2} = - 6 x