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Popular >

range of f(x)=2x^2+8x

Solution

range

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

Solution

f (x) \geq - 8

+1

Interval Notation

[- 8, \infty)

Solution steps

Vertex of

2 x^{2} + 8 x :

Minimum

(- 2, - 8)

For a parabola

a x^{2} + b x + c

with Vertex

(x_{v}, y_{v})

If

a < 0

the range is

f (x) \leq y_{v}

If

a > 0

the range is

f (x) \geq y_{v}

a = 2,

Vertex

(x_{v}, y_{v}) = (- 2, - 8)

f (x) \geq - 8

Popular Examples

domain of (\sqrt[3]{x-9})/(x^3-9)domain

inverse of e^{1+arcsin(x)}inverse

e^{1 + a r c s i n (x)}

symmetry x^3+5 symmetry

x^{3} + 5

inflection f(x)=x(x-9sqrt(x))inflection points

f (x) = x (x - 9 x)

perpendicular y=-2/3 x-3 perpendicular

y = - \frac{2}{3} x - 3

Frequently Asked Questions (FAQ)

What is the range of f(x)=2x^2+8x ?
The range of f(x)=2x^2+8x is f(x)>=-8