What is the vertices f(x)=x^2+16x+64 ?

The vertices f(x)=x^2+16x+64 is Minimum (-8,0)

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vertices f(x)=x^2+16x+64

vertices

f (x) = x^{2} + 16 x + 64

Minimum (- 8, 0)

Solution steps

y = x^{2} + 16 x + 64

The parabola parameters are:

a = 1, b = 16, c = 64

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} = 0

Therefore the parabola vertex is

(- 8, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 8, 0)

y = 2 x^{2} - 4 x + 5

y = 6 x^{2} - 1

(x + 5)^{2} = - 20 (y - 2)

7 x^{2} - 10 x + 5

f (x) = - 2 x^{2} + 80 x - 10

What is the vertices f(x)=x^2+16x+64 ?
The vertices f(x)=x^2+16x+64 is Minimum (-8,0)