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)

y = - 5 x + 7

f (x) = \frac{3 x - 4}{4 x + 9}

y = - \frac{1}{2} x^{2}

f (x) = 3 + x

\frac{x ^{2} + 5 x}{x ^{2} + 7 x + 10}