What is the range of f(x)=k(x)(x^{2+1})=x(x+6)+1 ?

The range of f(x)=k(x)(x^{2+1})=x(x+6)+1 is f(x)>=-8

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range of f(x)=k(x)(x^{2+1})=x(x+6)+1

range

f (x) = k (x) (x^{2 + 1}) = x (x + 6) + 1

f (x) \geq - 8

Interval Notation

[- 8, \infty)

Solution steps

Vertex of

x (x + 6) + 1 :

Minimum

(- 3, - 8)

For a parabola

a x^{2} + b x + c

with Vertex

(x_{v}, y_{v})

a < 0

the range is

f (x) \leq y_{v}

a > 0

the range is

f (x) \geq y_{v}

a = 1,

Vertex

(x_{v}, y_{v}) = (- 3, - 8)

f (x) \geq - 8

f (x) = 2 x - 48

(- 3) (- 11.9)

f (x) = x^{3} - 5 x^{2} - 8 x - 4

f (x) = \frac{2}{4 - 3 x + x ^{2}}

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

What is the range of f(x)=k(x)(x^{2+1})=x(x+6)+1 ?
The range of f(x)=k(x)(x^{2+1})=x(x+6)+1 is f(x)>=-8